A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate the maximum height to which it rises.
(ii) The total time it takes to return to the surface of the earth
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Answers
Answer:
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.
Explanation:
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Answer:
Total time T = Time to ascend (Ta) + Time to descend (Td)
V = u – gt
0 = 49 – 9.8 x Ta
Ta = (49/9.8) = 5s
Also, Td = 5s
Therefore T = Ta + Td
T = 5 + 5
T = 10s
Explanation:
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