the ball is thrown vertically upwards with a velocity of 39 m/s calculate the total time taken to return the surface of earth
Answers
Answer:
ANSWER
Given Initial velocity of ball, u=49 m/s
Let the maximum height reached and time taken to reach that height be H and t respectively.
Assumption: g=9.8 m/s
2
holds true (maximum height reached is small compared to the radius of earth)
Velocity of the ball at maximum height is zero, v=0
v
2
−u
2
=2aH
0−(49)
2
=2×(−9.8)×H
⟹H=122.5 m
v=u+at
0=49−9.8t
⟹t=5 s
∴ Total time taken by ball to return to the surface, T=2t=10 s
Explanation:
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Answer:
Explanation:
Given parameters
Initial velocity of the ball (u) = 49m/s.
The velocity of the ball at maximum height (v) = 0.
g = 9.8m/s2
Let us considered the time taken is t to reach the maximum height H.
Consider a formula,
2gH = v2 – u2
2 × (- 9.8) × H = 0 – (49)2
– 19.6 H = – 2401
H = 122.5 m
Now consider a formula,
v = u + g × t
0 = 49 + (- 9.8) × t
– 49 = – 9.8t
t = 5 sec
(1) The maximum height to ball rises = 122.5 m
(2) The total time ball takes to return to the surface of the earth = 5 + 5 = 10 sec.