If a stone is thrown up straight vertically with a speed of 12mls, what will be (a) maximum height it can reach 1) time taken to each maximum height c) time taken by it to come from maximum height back to initial position from when it was thrown. take g=10m/s^2
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it's phy the answer is 12
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Given parameters :-
Initial velocity of the ball (u) = 49m/s.
The velocity of the ball at maximum height (v) = 0.
g = 9.8m/s2
Find out
1) The maximum height to which it rises.
2) The total time it takes to return to the surface of the earth.
Solution
Let us consider the time 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
Answer
(1) The maximum height to which the ball rises = 122.5 m
(2) The total time ball takes to return to the surface of the earth = 5 + 5 = 10 sec.
hope it helps :)
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