Question

A projectile is fired from the top of a 120 m tall building with an initial velocity of Vi = 72.0 m/s and an initial angle theta i =25.0° above the horizontal. Answer the following questions.[Use g= 9.80 m/s^2]Question 12What is the distance traveled by the projectile in the horizontal direction?Round your answer to 3 significant figures.Question 13What is the speed of the projectile the moment it touches the ground?Round your answer to 3 significant figures.

Answer 1

Given data

(12)

*The given height of the building is h = 120 m

*The given initial velocity is v_i = 72.0 m/s

*The angle above the horizontal is

`\theta=25^0`

*The value of acceleration due to gravity is g = 9.80 m/^2

The initial horizontal velocity of the projectile is calculated as

`\begin{gathered} v_x_{}=v_i\cos \theta \\ =(72.0)\cos 25^0 \\ =65.25\text{ m/s} \end{gathered}`

The initial vertical component velocity of the projectile is calculated as

`\begin{gathered} v_y_{}_{}=v_i\sin \theta \\ =(72.0)\sin 25^0 \\ =30.42\text{ m/s} \end{gathered}`

The time taken by the projectile is calculated by the kinematic equation of motion as

`H=v_yt+(1)/(2)gt^2`

Treating the vertically upward direction to be a negative and the vertically downward direction to be positive.

Substitute the known values in the above expression as

`\begin{gathered} 120=-(30.42)t+(1)/(2)(9.8)t^2 \\ t=8.95\text{ s} \end{gathered}`

The expression for the distance traveled by the projectile is given as

`D=v_x* t`

Substitute the known values in the above expression as

`\begin{gathered} D=(65.25)*(8.94) \\ =583.33\text{ m} \\ =584\text{ m} \end{gathered}`