A projectile is shot from the edge of a cliff h = 125 m above ground level with an initial speed of v0 = 115 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-39

Question

(a) Determine the time taken by the projectile to hit point P at ground level.
(b) Determine the range X of the projectile as measured from the base of the cliff.
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
(d) What is the the magnitude of the velocity?
(e) What is the angle made by the velocity vector with the horizontal?

Answer 1

We don't get to see the figure 3-39, but I assume that the angle 37° is an angle of elevation, that is, it is shot upwards.

Resolve the velocity into the horizontal (cos37°) and vertical (sin37°) components.

Without air resistance, the horizontal component will not vary, i.e. it is a constant velocity.

The vertical component can be treated like a projectile shot vertically upwards, and the usual kinematics formulae will apply.
v0=115sin37° m/s
a=-9.8 m/s/s
S= -125 m when it hits the ground
t=time in seconds
2aS = v0t+(1/2)at²
Solve for the quadratic to get t.
Can you take it from here?

Answer 2

stuff

Answer 3

y 2aS = v0t+(1/2)at² ?????

Answer 4

I need help with the same problem! :(