a ball is thrown vertically upward with a velocity of 49m/s . calculate a) maximum height to which it rises b)total time it takes to return to the surface of the earth ?
Answers
Answered by
4
Explanation:
u = 49m/s
v = 0 m/s
( because at height point velocity is 0)
a = -g = (-10)
First let's calculate time
v = u+at
0 = 49 + (-10)t
t = 4.9 sec
a) maximum height
S = ut +1/2 at^2
S = 240.1 (-5) (4.9)
S = 215.6
b)total time it takes to return to the surface of the earth
Time of acent = time of decent
2×4.9
9.8 sec
Hope it helps uh'!
Answered by
7
Answer :
Given -
- u = 0 m/s
- v = 49 m/s
- g = -9.8 m/s²
To Find -
- s ?
Solution -
We know that, v² - u² = 2gs.
Hence,
⇒ 49² - 0² = 2*-9.8*s
⇒ 2401 = 19.6*s
⇒ s = 2401/19.6
⇒ s = 122.5 m
Hence, the Height which the ball rises is 122.5 m.
Now, We know that, s = ut + ½gt²
⇒ 122.5 = 0*t + ½*9.8*t²
⇒ 122.5 = 4.9t²
⇒ t² = 122.5/4.9
⇒ t² = 25
⇒ t = 5 seconds
Hence, the total time it takes to return to the surface to the earth is 5 seconds.
Equations of Motion :
- v = u + at
- s = ut + ½gt²
- v² - u² = 2gs
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