A person who is 1 m tall throws a ball with a speed of 29 m/s at an angle of 57

Question

A person who is 1 m tall throws a ball with a speed of 29 m/s at an angle of 57 degrees above the horizontal, as shown in the diagram below. h = 1 m, v = 29 m/s, = 57 degrees Use the acceleration due to gravity as g = 10 m/s2.

What are the x and y components of the ball's velocity when it is at its maximum height? ( , ) m/s

What are the x and y components of the ball's velocity the instant before it hits the ground? ( , ) m/s

What is the ball's speed (the magnitude of the velocity) the instant before it hits the ground? m/s

Explanation / Answer

a)

here

at max height vy = 0

vx = 29 * cos57 = 15.8 m/s

b)

by using the second equation of motion

s = u*t+ 0.5 * a*t^2

-1 = 29 * sin57 * t + 0.5 * -9.8 * t^2

24.32t - 4.9t^2 + 1 = 0

by solving this quadratic equation we get

t = 5 sec

the x , y component before the ball hit the ground

vy = 29 * sin57 - 9.8 * 5 = -24.68 m/s

vx = 29 * cos57 = 15.8 m/s

c)
the magnitude is

v = sqrt(15.8^2 + 24.68^2) = 29.3 m/s

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