A projectile is shot from the edge of a cliff h = 165 m above ground level with

Question

A projectile is shot from the edge of a cliff h = 165 m above ground level with an initial speed of v0 = 105 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-35.


Figure 3-35
(a) Determine the time taken by the projectile to hit point P at ground level.
Your answer is correct. s
(b) Determine the range X of the projectile as measured from the base of the cliff.
Your answer is correct. km
(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.)
horizontal Your answer is incorrect. m/s
vertical Your answer is correct. m/s
(d) What is the the magnitude of the velocity?
m/s
(e) What is the angle made by the velocity vector with the horizontal?
° (below the horizontal)
(f) Find the maximum height above the cliff top reached by the projectile.
m

Explanation / Answer

(a) You did not provide your answer for part a, so I will just work out briefly here:

s = ut + 0.5at^2
-165 = (105sin37)t - 0.5(g)(t^2)
t = 12.3sec or t = -2.23sec (rejected because time has no negative value)

(c) For part C, the horizontal component will be the same as the initial horizontal component: 105cos37 = 83.9m/s For vertical component, use the law of kinematics: v^2 = u^2 + 2as v^2 = (105sin37)^2 + 2(-g)(-165)               (Taking upwards as positive) v = 85.03m/s (d) The magnitude is referring to the resultant velocity that both the horizontal and vertical components are contributing to: Vfinal^2 = (85.03)^2 + (83.9)^2 Vfinal = 119m/s (e) Similarly for the angle: tan? = 85.03/83.9 ? = 45.4 degrees (f) The maximum height reached is when the vertical component of the velocity reaches 0m/s. Thus, v^2 = u^2 + 2as v = 0, 0 = (105sin37)^2 + 2(-g)(-s) s = 204m

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