125k views
5 votes
What is the full distance when an object is thrown at 35 m/s at an angle of 45 degrees

What is the full distance when an object is thrown at 35 m/s at an angle of 45 degrees-example-1

2 Answers

3 votes

Answer:

To calculate the full distance traveled by an object thrown at a velocity of 35 m/s at an angle of 45 degrees, we need to consider the horizontal and vertical components of the motion separately.

The horizontal component of the motion remains constant throughout the trajectory and is given by:

Horizontal distance = (Initial velocity) * (Time of flight) * cos(angle)

In this case, the initial velocity is 35 m/s, the angle is 45 degrees, and we need to find the time of flight.

The time of flight can be calculated using the vertical component of the motion. The vertical motion can be described using the equation:

Vertical displacement = (Initial velocity * sin(angle))^2 / (2 * acceleration)

Where the initial velocity is 35 m/s, the angle is 45 degrees, and the acceleration is the acceleration due to gravity, approximately 9.8 m/s^2.

The vertical displacement is zero at the highest point of the trajectory since the object comes back down to the same height it was launched from. So we can solve the equation for the time of flight.

Using these calculations, we can find the horizontal distance traveled by the object.

Let's calculate step by step:

Step 1: Calculate the time of flight

Vertical displacement = 0 (at the highest point)

0 = (35 * sin(45))^2 / (2 * 9.8)

0 = (24.75^2) / 19.6

0 = 616.0125 / 19.6

0 = 31.43

Step 2: Calculate the time of flight

Vertical displacement = (Initial velocity * sin(angle)) * time - (1/2) * acceleration * time^2

0 = (35 * sin(45)) * time - (1/2) * 9.8 * time^2

0 = 24.75 * time - 4.9 * time^2

4.9 * time^2 - 24.75 * time = 0

time * (4.9 * time - 24.75) = 0

time = 0 (initial point) or 24.75 / 4.9

time = 5.05 seconds

Step 3: Calculate the horizontal distance

Horizontal distance = (Initial velocity) * (Time of flight) * cos(angle)

Horizontal distance = 35 * 5.05 * cos(45)

Horizontal distance = 35 * 5.05 * (sqrt(2)/2)

Horizontal distance = 88.96 meters

Therefore, when an object is thrown at 35 m/s at an angle of 45 degrees, the full distance traveled is approximately 88.96 meters.

User Alex Fragotsis
by
8.4k points
4 votes

Okay, here are the steps to calculate the full distance traveled when an object is thrown at a certain speed and angle:

You have the initial velocity (v): 35 m/s

You have the launch angle (θ): 45 degrees

We need to split the initial velocity into its horizontal (vx) and vertical (vy) components.

To calculate vx (horizontal component):

vx = v * cosθ

vx = 35 * cos(45) = 24.7 m/s

To calculate vy (vertical component):

vy = v * sinθ

vy = 35 * sin(45) = 24.7 m/s

We can calculate the horizontal distance (d) traveled using:

d = vx * t (where t is time)

Since there is no air resistance, the vertical velocity (vy) will remain constant. This means the time the object is in the air is:

t = vy / g (where g is acceleration due to gravity, 9.8 m/s^2)

t = 24.7 / 9.8 = 2.52 seconds

Now we can calculate the full horizontal distance traveled:

d = vx * t

d = 24.7 * 2.52

= 62.3 meters

So the full distance the object will travel when thrown at 35 m/s at a 45 degree angle is approximately 62 meters.

Let me know if you have any other questions!

User Null Canvas
by
8.0k points

No related questions found