Question

a ball thrown up is caught by thrower after 4 seconds.how high did it go and with what velocity was it thrown?how far below it highest point was in 3 seconds after start?Acceleration due to gravity is 9.8m/s2

Answer 1

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Answer 2

t = 4sec
v = 0
g = 9.8m/s^2

v = u + gt
0 = u -9.8×4
u = 39.2 m/s

v^2-u^2 = 2gh
0 - (39.2)^2 = 2×(-9.8)×h
h = 78.4 m

u = 39.2 m/s
t = 3
g = 9.8m/s^3

h' = ut+(1/2)gt^2
= 39.2×3 -(4.9×9)
= 117.6 - 44.1
h' = 73.5
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•The ball goes 78.4 m high
• The velocity with which it was thrown is 39.2m/s
• The far below it's highest point the ball was in 3 after start = h-h' = 78.4-73.5 = 4.9 m
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