A projectile is shot from the edge of a cliff h= 245 m above ground level with a

Question

A projectile is shot from the edge of a cliff h= 245 m above ground level with an initial speed of v0= 155 m/s at an angle of 37.0 with the horizontal, as shown in the Figure.

a. What are the horizontal and vertical components of the initial velocity?

b. Determine the time taken by the projectile to hit point P at ground level

c. Determine the range X of the projectile as measured from the base of the cliff

d. Find the maximum height above the cliff top reached by the projectile

e. What is its speed right before it hits the ground?

Please Explain!! how to solve each step.!

Explanation / Answer

a) horizontal and vertical components of the initial velocity
The initial vertical velocity Vv= 155sin37 = 93.28 m/s and the horizontal initial velocity Vh =155cos37 = 123.79 m/s
b) Using: Y = Yo - Vv t + (1/2)gt^2 <=== Setting down as positive y-direction
Substituting;
245 = 0 - 93.28 t + (4.9) t^2

Rearranging into standard quadratic form:
4.9t^2 - 93.28t - 245 = 0

Use quadratic formula to solve for t (a= 4.9; b= -93.28; c= -245)
t = [-b +/- sqrt(b^2 - 4ac)]/(2a)
t = [93.28 +/- sqrt(13503.1584)]/9.8
t = [93.28 +/- 116.2031]/9.8 <=== Disregard negative root as it is meaningless
t = 21.376 seconds

c) Using: X = Xo + Vh t + (1/2)at^2, and since there is no horizontal acceleration

Substituting:
x = 0 + (155 cos 37)(21.376) + 0 =123.79 x 21.376
x = 2646.14 m

d) Maximum height when Vy = 0
Using: V^2 = Vh^2 + 2ad
(0)^2 = (155 sin 37)^2 + 2(-9.8)d
d = 443.94 m

e)V^2 = Vh^2 + (Vv -gt)^2 where Vh=93.28 m/s and Vv = 123.79m/s, t=21.376 s
V = 126.66 m/s

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