Question

Aman tosses a dart upward with a velocity of 14.1 m/s at 60° angle.

How much time is it aloft?
What is its max height and range?

Answer 1

We are given:

Angle of projection of the dart = 60°

initial velocity of the dart = 14.1 m/s

Horizontal and vertical component of the Dart:

Using basic trigonometry, we can see that:

Vertical component of the Dart = 14.1*Sin 60° = 12.2 m/s

Horizontal component of the Dart = 14.1*Cos 60° = 7.05 m/s

Time the Dart is Aloft:

Explaining the concept:

Once the dart is in mid-air, there will be no force that will reduce its horizontal velocity

So, the dart will be moving at a constant horizontal velocity of 7.05 m/s

but the force of gravity will be applied on the dart while mid-air and will pull it downwards with an acceleration of 9.8 m/s²

Hence, the time the dart will be aloft is the same as the time taken by the dart to reach the ground if we threw it vertically upwards at a velocity of 12.2 m/s

Solving for the time taken for the dart to reach the ground:

initial velocity = 12.2 m/s

velocity at max height = 0 m/s

acceleration = -9.8 m/s

time taken to reach max height:

v = u + at using the first equation of motion

replacing the variables

0 = 12.2 + (-9.8)t

t = -12.2 / -9.8

t = 1.24 seconds

Time taken to reach the ground:

time taken to reach the ground = 2 * time taken to reach max height

(Since the dart will take the same time to come back down)

Time taken to reach the ground = 2 * 1.24

Time taken to reach the ground = 2.48 seconds

Therefore, the dart will hit the ground and stop 2.48 seconds after throwing

Max height of the Dart:

Explaining the concept:

From newton's second equation of motion

s = ut + 1/2 at²

Since we need the max vertical height, we need the vertical component of the values,

so this equation can be rewritten for vertical height as :

s(vertical) = u(vertical)* t + 1/2 * a(vertical) * t²

from the last section, we know that the dart reaches its maximum height at t = 1.24 seconds

replacing the values in this equation:

s(vertical) = (12.2 * 1.24) + 1/2 * (-9.8) * (1.24)²

s(vertical) = 15.128 + (-7.5)

s(vertical) = 7.63 m

Therefore, the maximum height of the dart is 7.63 m

The range of the Dart:

Explaining the concept:

The range of the dart is the horizontal distance covered by the dart

from the second section, we know that the dart travels horizontally with a constant velocity of 7.05 m/s

from the third section, we know that the total time taken by the dart to hit the ground is 2.48 seconds

Since the horizontal velocity of the dart is constant, we can say that it moved horizontally for 2.48 seconds at a constant velocity of 7.05 m/s

Solving for the range:

s = ut + 1/2 at² From the second equation of motion

this equation can be rewritten for horizontal distance as:

s(horizontal) = u(horizontal)* t + 1/2 (a)(t)²

The dart is moving at a constant velocity, that means that its acceleration is 0

s(horizontal) = u(horizontal)* t + 1/2 (0)(t)²

s(horizontal) = u(horizontal)* t

replacing the variables

s(horizontal) = 7.05 * 2.48 [here, 2.48 is the time taken by the dart to reach the ground and stop]

s(horizontal) = 17.484 m

The horizontal distance covered by the dart is 17.484 m

Therefore, the range of the dart is 17.484 m

Answer 2

time = 2.49 s

max height = 7.61 m

range = 17.57 m

See below for exact values.

Explanation:

Vertical/Horizontal Components:

Since we are given the angle at which the object is tossed, this object is in projectile motion.

We are given the initial velocity of the dart, and we can use the angle to solve for its horizontal and vertical components.

Horizontal component:

• 
  `(v_i)_x =v_i \cdot cos(\theta)`
• 
  `(v_i)_x=14.1\cdot cos(60)`

Vertical component:

• 
  `(v_i)_y =v_i \cdot sin(\theta)`
• 
  `(v_i)_y=14.1 \cdot sin(60)`

Time In The Air:

In order to find the time that the dart stays in the air, we can use the constant acceleration equation that does not include displacement. This equation is:

• 
  `v_f=v_i+at`

Since we solve for time using the vertical motion of the projectile, we will use this equation in terms of the y-direction.

• 
  `(v_f)_y = (v_i)_y+a_yt`

We don't know what the final velocity of the dart when it reaches the ground is, but we do know that its final velocity when it reaches its maximum height is 0 m/s.

Therefore, we can solve for half of the full time that it takes the object to reach the ground, and double this value at the end.

Let's set the downwards direction to be negative and the upwards direction to be positive.

Using our knowledge and previous calculations, we know that:

• 
  `(v_f)_y = 0`
• 
  `(v_i)_y=14.1\cdot sin(60)`
• 
  `a_y=-9.8`

The acceleration of gravity is in the y-direction and facing downwards, so that's why it is -9.8 m/s².

Substitute these values into the constant acceleration equation.

• 
  `0=14.1\cdot sin(60) + (-9.8)t`

Subtract 14.1 * sin(60) from both sides of the equation.

• 
  `-14.1\cdot sin(60)=-9.8t`

Divide both sides of the equation by -9.8 to solve for t.

• 
  `\displaystyle{(-14.1\cdot sin(60))/(-9.8) =t}`

• 
  `t=1.246016142`

Now, this is only the time for half of the object's trajectory. Double it to find the total time the dart is aloft:

• 
  `2t = 2.492032284`

The dart is in the air for a total of 2.49 seconds.

Maximum Height:

We can find the maximum height of the dart by using another constant acceleration equation.

Since we don't have the final velocity of the object, we can use this equation:

• 
  `x_f=x_i+v_it+(1)/(2)at^2`

Subtract
`x_i` from both sides to get the change in position, or delta x.

• 
  `\triangle x=v_it+(1)/(2)at^2`

In order to find the maximum height, we need to use this equation in terms of the y-direction.

• 
  `\triangle x_y=(v_i)_yt+(1)/(2)a_yt^2`

Remember that time is the same regardless of the x- or y- direction.

Now, we can solve for the displacement in the y-direction by plugging in the values that we know.

• 
  `(v_i)_y=14.1\cdot sin(60)`
• 
  `t=1.246016142`
• 
  `a_y=-9.8`

Note that we are using half of the time t, since this is where the maximum height occurs.

Plug these known values into the constant acceleration equation:

• 
  `\triangle x_y=[14.1 \cdot sin(60)] (1.246016142) + (1)/(2)(-9.8)(1.246016142)^2`

• 
  `\triangle x_y=(15.21505102) + (-4.9)(1.552556226)`

• 
  `\triangle x_y=15.21505102-7.607525508`

• 
  `\triangle x_y=7.60752551`

The maximum height of the dart is 7.61 meters.

Range:

Finding the range of the object in projectile motion involves the same constant acceleration equation, but this time we are solving for the displacement in the horizontal direction (x-direction). Therefore:

• 
  `\triangle x_x=(v_i)_xt+(1)/(2)a_xt^2`

We know the horizontal component of the initial velocity vector, which is what we will be using.

The acceleration, of an object in projectile motion, in the x-direction is always 0 m/s².

We will use the full time of the object since we want to find the entire horizontal distance that the object covers. We have:

• 
  `(v_i)_x=14.1\cdot cos(60)`
• 
  `t=2.492032284`
• 
  `a_x=0`

Plug these values into the constant acceleration equation.

• 
  `\triangle x=[14.1\cdot cos(60)](2.492032284)+(1)/(2) (0)(2.492032284)^2`
• 
  `\triangle x=[14.1\cdot cos(60)](2.492032284)`
• 
  `\triangle x=17.5688276`

The range of the dart is 17.57 meters.