An elevated catapult launches a ball with a velocity of 30.0 m/s, directed at an

Question

An elevated catapult launches a ball with a velocity of 30.0 m/s, directed at an angle of 54 degree above the horizontal. The catapult is elevated vertically 13.0 m above the ground. (a) What is the horizontal component of the ball's velocity when it launches? m/s (b) What is the vertical component of the ball's launch velocity? m/s (c) When does the ball land? m/s (d) How far away does the ball land, as measured horizontally from the ground under the catapult? m (e) At what time does the ball reach its highest point? s (f) When the ball is at its highest point, how far is it, as measured horizontally from the catapult? m (g) When the ball is at its highest point, how high is it relative to the ground? m

Explanation / Answer

u = 30 m/s

theta = 54 degree

h = 13 m

a) horizontal component of velocity = u * cos(theta)

horizontal component of velocity = 30 * cos(54 degree)

horizontal component of velocity = 17.6 m/s

b)

at the launch

vertical component of velocity = u * sin(theta)

vertical component of velocity = 30 * sin(54 degree)

vertical component of velocity = 24.3 m/s

c)

let the time of flight is t

-13 = 24.3 * t - 0.50 * 9.8 * t^2

solving for t

t = 5.44 s

the time of flight is 5.44 s

d)

horizontal distance = 5.44 * 17.6

horizontal distance = 95.7 m

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