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 a) What are the horizontal and vertical components of the initial velocity? b) Determine the time taken by the projectile to hit the ground. 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? Answers are A) 124 and 93.3 B) 21.4s C) 2.65 km D)444m E) idk Please help me understand how to work these problems

Explanation / Answer

Here ,

h = 245 m

v0 = 155 m/s

theta = 37 degree

a)

horizontal component of velocity = v0 * cos(theta)

horizontal component of velocity = 155 * cos(37)

horizontal component of velocity = 123.7 m/s

vertical component of velocity = v0 * sin(theta)

vertical component of velocity = 155 * sin(37)

vertical component of velocity = 93.3 m/s

part b)

let the time of flight is t

Using second equation of motion

y = u * sin(theta) * t + 0.5 * g * t^2

-245 = 93.3 * t - 0.5 * 9.8 * t^2

solving for t

t = 21.4 s

the time taken to reach ground is 21.4 s

c)

range of the projectile = 21.4 * 123.7 m

range of the projectile = 2647.2 m

d)

for the maximum height above cliff

maximum height above cliff = (vy^2)/2g

maximum height above cliff = 93.3^2/(2 * 9.8)

maximum height above cliff = 444.13 m

the maximum height above cliff is 444.13 m

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