A ball is thrown upwards with a velocity 20 ms from the top of tower having a height 50 m. whai is time elasped when it passes again edge of tower
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A ball is thrown upwards with a velocity 20 ms from the top of tower having a height 50 m. whai is time elasped when it passes again edge of tower
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Answer:
Time taken to pass edge of tower
= 2 seconds
Step by step explanations :
given that,
a ball is thrown upwards with a velocity 20 m/s from the top of tower
here,
initial velocity of the ball = 20 m/s
let the maximum height reached by ball from the initial position be h
so,
at its maximum height velocity will 0
so,
final velocity of the ball = 0 m/s
now we have,
Initial velocity(u) = 20 m/s
final velocity(v) = 0 m/s
gravitational acceleration(g) = -10 m/s²
Now,
by the gravitational equation of motion,
v = u + gt
where,
t is the time taken to reach maximum height
putting the values,
0 = 20 + 2(-10)t
-20t = -20
t = -20/-20
t = 1 s
so,
time taken to reach maximum height
= 1 seconds
now we know that,
Time taken to reach maximum height
= time taken to reach its initial position
so,
time elapsed when it passes again edge of tower = 1 + 1
= 2 seconds
Time taken to pass edge of tower
Time taken to pass edge of tower= 2 seconds
Step-by-step explanation:
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