Question

A large catapult is used to launch a pumpkin. When the catapult releases the pumpkin, the pumpkin is located 25.0 m above the ground. If catapult

launches the pumpkin at a velocity of 25 m/s at 20 degrees above the horizontal and the pumpkin lands on the ground
(continued from
previous problem)
b. How far will it land from the catapult?
c. What is the maximum height it will reach in relation to the ground?

Answer 1

Hi there!

Part C.

We can begin by solving for Part C first, which will allow us to do part B next.

Use the kinematic equation:

`v_(fy)^2 = v_(iy)^2 + 2ad_y`

At the top of the trajectory, the final VERTICAL velocity is 0 m/s, so:

`0 = v_(iy)^2 + 2ad_y`

Solve for the initial vertical velocity by finding the vertical component of the given velocity:

`v_y = vsin\theta = 25sin(20) = 8.55 m/s`

Plug in the acceleration due to gravity:

`0 = (8.55^2) + 2(-9.8)d\\\\19.6d = 73.11\\\\d = 3.73 m`

Add this to the initial height to get the total max height:

`h_(max) = 25 + 3.73 = \boxed{28.73 m}`

Part B.

We must solve for the times taken for the pumpkin to reach its max height and to fall down.

The time taken to reach the max height can be solved by:

`t = (\Delta v)/(a)\\\\t = (v_y)/(g) = (8.55)/(9.8) = .873 s`

Now, solve for the time to fall from the max height solved above using the rearranged kinematic equation:

`t = \sqrt{(2h)/(g)} = \sqrt{(2(28.73))/(9.8)} = 2.42s`

Add the times:

`2.42 + .873 = 3.293 s`

Now, using the distance/speed/time equation:

`d_x = v_xt\\\\v_x = vcos\theta = 25cos(20) = 10.20 m/s\\\\d = 3.293(10.20) = \boxed{33.6 m}`