Question

A projectile is shot from the edge of a cliff 115 mm above ground level with an initial speed of v0v0 = 65 m/sm/s at an angle of 35.0∘∘ with the horizontal, as shown in the figure (Figure 1). Figure 1 of 1

Part A Determine the time taken by the projectile to hit point PP at ground level. Express your answer to three significant figures and include the appropriate units. tt = nothingnothing SubmitRequest Answer Part B Determine the distance XX of point PP from the base of the vertical cliff. Express your answer to three significant figures and include the appropriate units. XX = nothingnothing SubmitRequest Answer Part C At the instant just before the projectile hits point PP, find the horizontal and the vertical components of its velocity. Express your answers using three significant figures separated by a comma. vxvx, vyvy = nothing m/sm/s SubmitRequest Answer Part D At the instant just before the projectile hits point PP, find the magnitude of the velocity. Express your answer to three significant figures and include the appropriate units. vv = nothingnothing SubmitRequest Answer Part E At the instant just before the projectile hits point PP, find the angle made by the velocity vector with the horizontal. Express your answer to three significant figures and include the appropriate units. θθ = nothingnothing below the horizon SubmitRequest Answer Part F Find the maximum height above the cliff top reached by the projectile.

Answer 1

Given the height of the cliff h = 115m, the initial speed of projection v0 = 65m/s and the angle of projection . The path of the projectile is shown in the figure below.

A) The time taken by the projectile to go from A to B is the time of flight T of the projectile. It is given by,

The time taken from B to P is given by,

We take 2.35s.

The total time taken is,

So the time taken by the projectile to hit point P at ground level is 9.96s.

B) The diatance X is the sum of the horizontal distances from A to B and B to P. The horizontal distance from A to B is the range of the projectile R. It is given by,

The The horizontal distance from B to P is the distance travelled with the velocity . It is give by,

The total distance is given by,

So the distance X of point P from the base of the vertical cliff is 530m.

C) The horizontal component of velocity of the projectile just before it hits P is given by,

The vertical component of velocity of the projectile just before it hits P is given by,

Therefore,

D) The magnitude of velocity of the projectile just before it hits P is given by,

So the magnitude of velocity of the projectile just before it hits P is 80.4m/s.

E) The direction of the final velocity is,

So the angle made by the velocity vector with the horizontal just before the projectile hits the ground is .

F) The height above the cliff top reached by the projectile is the maximum height of the projectile. It is given by,

So the maximum height above the cliff top reached by the projectile is 70.9m.