A projectile is shot from the edge of a cliff h = 265 m above ground level with

Question

A projectile is shot from the edge of a cliff h = 265 m above ground level with an initial speed of v0 = 125 m/s at an angle of 37.0 degree with the horizontal, as shown in the figure below. (a) Determine the time taken by the projectile to hit point P at ground level. ____ s (b) Determine the range X of the projectile as measured from the base of the cliff. _____ km (c) At the instant just before the projectile hits point P, find the horizontal and vertical components of its velocity. (Take up and to the right as positive directions.) horizontal ____________ m/s vertical ____________ m/s (d) What is the magnitude of the velocity? _________ m/s (e) What is the angle made by the velocity vector with the horizontal? _____ degree (below the horizontal) (f) Find the maximum height above the cliff top reached by the projectile. _________ m

Explanation / Answer

1) Since it is launched with an angle of 37 deg, it has two components - Vertical and horizontal;
The horizontal velocity will remain constant and only vertical velocity will vary because of the acceleration due to earth's gravitational pull, which we denote by g = 9.8 m/sec^2

2) Initial vertical velocity Vu = 125 x sin37 = 85.23 m/sec
Horizontal Velocity H = 125 x cos37 = 99.8 m/sec [this is constant]

3) Time taken to reach max height, that is vertical velocity is 0;
Applying, v = u - gt, 0 = 85.23 - 9.8t; ==> t = 8.7 seconds
It takes same 8.7 seconds to again reach back the top of the cliff.
==> Time taken from edge of cliff - back to edge of cliff = 17.4 seconds

4) Velocity just before hitting the ground:
Using, v^2 = u^2 + 2gh, v^2 = (85.23)^2 + 2 x 9.8 x265 [From 2, Uv = 85.23; h = 265 data; g is positive, since now it is considered down from cliff edge]
==> v = 111.6 m/sec

5) time taken from cliff edge to P: t = (111.6 - 85.23)/9.8 = 2.7 seconds

6) Total time taken to reach P = 17.4 + 2.7 = 20.1 seconds

7) Range X = Horizontal velocity x total time taken = 99.8 x 20.1 = 2006 m,
say nearly 2 KM

8) Just before hitting the magnitude of velocity = [sqrt{(95.135)^2 + (107.816)^2}] = 143.788 m/s

9) Angle, cos(<A) = Horizontal velocity/Velocity just before hitting the ground = 107.816/143.788
==> <A = 41.42 deg

10) Max height: Using v^2 = u^2 - 2gh; [u, initial vertical velocity = 81.245 m/s; v = 0; g = 9.8]
==> h = (81.245)^2/19.6 = 336.773 m

ANSWERS: (Answers rounded off to one decimal point)

a) Total time taken to hit the ground at P = 18 seconds.

b) Range X = 1.9 km

c) At the instant before hitting the ground,
the horizontal velocity = 107.8 m/sec
the vertical velocity = 95.1 m/sec

d) Magnitude of velocity just before hitting P = 143.8 m/s

e) The angle made with horizontal = 41.42 deg

f) The maximum height reached above the cliff top = 336.8 m

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