Question

if a projectile is launched upward with a speed of 7.44 m/s, and at an angle of 40.1° above horizontal, what is the hang time of this projectile?

Answer 1

The hang time of a projectile is determined by its initial vertical velocity, which is calculated from the initial speed and the angle of launch, and gravity. The time in the air is double the ascent time, calculated by dividing the vertical velocity by gravity and then multiplying by two.

Step-by-step explanation:

The hang time of a projectile launched at an angle can be calculated using the vertical component of the initial velocity and the acceleration due to gravity.

The vertical component (vy) is found by multiplying the initial speed by the sine of the launch angle: vy = v * sin(θ). The hang time (t) is then calculated as t = (2 * vy)/g, where g is the acceleration due to gravity (9.81 m/s2).

For a projectile launched upward with a speed of 7.44 m/s at an angle of 40.1° above the horizontal, the vertical component of the velocity is vy = 7.44 m/s * sin(40.1°).

After finding vy, we calculate the hang time using the formula t = (2 * vy)/g. This will give us the total time the projectile is in the air until it returns to the same vertical position from which it was launched.

The complete question is: if a projectile is launched upward with a speed of 7.44 m/s, and at an angle of 40.1° above horizontal, what is the hang time of this projectile? is:

Answer 2

So, the hang time of the projectile is approximately 0.95 seconds.

The hang time of a projectile is the total time it remains in the air. In order to find the hang time, we can use the following kinematic equation for the vertical motion of a projectile:

`\[ y = v_0 \sin(\theta) t - (1)/(2) g t^2 \]`

where:

-
`\( y \)` is the vertical displacement (which is zero at the peak of the trajectory),

-
`\( v_0 \)` is the initial velocity (7.44 m/s),

-
`\( \theta \)` is the launch angle (40.1°),

-
`\( t \)` is the time of flight,

-
`\( g \)` is the acceleration due to gravity (approximately 9.8 m/s²).

At the peak of the trajectory, the vertical displacement
`\( y \)` is zero. Therefore, we can set
`\( y = 0 \)` and solve for
`\( t \)`.

`\[ 0 = v_0 \sin(\theta) t - (1)/(2) g t^2 \]`

Rearrange the equation:

`\[ (1)/(2) g t^2 = v_0 \sin(\theta) t \]`

Now, solve for \( t \):

`\[ t = (2v_0 \sin(\theta))/(g) \]`

Substitute the known values:

`\[ t = \frac{2 * 7.44 \, \text{m/s} * \sin(40.1°)}{9.8 \, \text{m/s}^2} \]`

Calculate this expression to find the hang time of the projectile.

Let's continue solving for the hang time (\(t\)):

`\[ t = \frac{2 * 7.44 \, \text{m/s} * \sin(40.1°)}{9.8 \, \text{m/s}^2} \]`

`\[ t = (2 * 7.44 * \sin(40.1°))/(9.8) \]`

`\[ t \approx (2 * 7.44 * 0.6428)/(9.8) \]`

`\[ t \approx (9.29)/(9.8) \]`

`\[ t \approx 0.95 \, \text{s} \]`