A ball is thrown vertically upwards and returns to the thrower after 12 seconds find initial velocity and height attained
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A ball is thrown vertically upwards and returns to the thrower in 20 seconds. How would you calculate the velocity with which it was thrown and the maximum height attained by the ball, given g=10m/s2?
Once the ball is thrown, the only force acting on it is gravity, which means that it's acceleration is -9.81 m/s² (negative means downward). The trajectory of the ball's movement is symmetrical. So the magnitude of the initial velocity is the same as the final velocity, but in the opposite direction. The maximum height of the ball occurs halfway through its flight, in this case 10 s.
List the known and unknown quantities from the question.
u = initial velocity = ? m/s
v = final velocity = 0 m/s
g = acceleration due to gravity = -9.8 m/s²
t = time interval = 10 s
s = displacement (maximum height) = ? m
Calculate the initial velocity using the following kinematic equation.
v = u + gt
Solve for u.
u = v - gt
u = 0 - (-10 m/s² × 10 s) = 100 m/s
Calculate the maximum height using the following kinematic equation.
s = ½(v +u)t
s = 0.5 × (0 + 100 m/s) × 10 s = 500 m
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Answer:
Question

The ball returns to the ground after 6 seconds.
Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s
Let the velocity with which it is thrown up be u
(a). For upward motion,
v=u+at
∴ 0=u+(−10)×3
⟹u=30 m/s
(b). The maximum height reached by the ball
h=ut+21at2
h=30×3+21(−10)×32
h=45 m
(c). After 3 second, it starts to fall down.
Let the distance by which it fall in 1 s be d
d=0+21at′2 where t′=1 s
d=21×10×(1)2 =5 m
∴ Its height above the ground, h′=45−5=40 m
Hence after 4 s, the ball is at a height of 40 m above the ground.
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