Question

A ball is shot out at 25 degrees above horizontal by a slingshot at the edge of a cliff. If the initial speed is 65m/s and the height of the cliff is 150m.

a) how long does it take for the ball to hit the water?
b) What is the velocity when the ball is 24m below the cliff?

Answer 1

To answer part a, we use the equation for vertical displacement in projectile motion. For part b, we can use the same equation to find the time it takes for the ball to reach a certain displacement below the cliff.

Step-by-step explanation:

To answer part a of the question, we can use the equation for the vertical displacement of an object in projectile motion. The equation is:

Δy = V₀yt + 1/2gt^2

where Δy is the vertical displacement, V₀y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.

In this case, V₀y = V₀sin(θ) = 65sin(25) and Δy = -150m (negative because the height is decreasing).

Using this information, we can solve for t.

For part b, we can use the same equation to find the time it takes for the ball to reach a displacement of -24m (where -24m is the difference between the initial height of the cliff and the displacement of the ball below the cliff).

By setting Δy = -24m, we can solve for t.