#4) Use the below figure for projectile motion. You are working on your jump sho
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Question
#4) Use the below figure for projectile motion. You are working on your jump shot. The magnitude of the initial velocity for the ball is vo 8.0 m/s and the angle you launch it is at 32.0. Note the ball's initial starting height is h 1.70m. Most of your points will come from setting up the equations correctly and your algebra manipulation prior to plugging in numbers, Numbers last. (A) Calculate the maximum height the ball goes relative to the ground. (B) Calculate the total time the ball is in the air, from h to the time it hits the ground. (C) The maximum horizontal distance the person can throw ball given these conditions.Explanation / Answer
here,
4)
initial speed , u = 8 m/s
theta = 32 degree
initial height , h0 = 1.7 m
a)
the maximum height , h = h0 + (u * sin(theta))^2 /2g
h = 1.7 + (8 * sin(32))^2 /( 2 * 9.81) m
h = 2.62 m
b)
let the time taken to reach the ground be t
h0 = - u * sin(theta) * t + 0.5 * g * t^2
1.7 = - 8 * sin(32) * t + 0.5 * 9.81 * t^2
solving for t
t = 1.16 s
c)
the maximum horizontal distance , x= u * cos(theta) * t
x = 8 * cos(32) * 1.16 m
x = 7.87 m
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