Question 13 A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises. (ii)the total time it takes to return to the surface of the earth.
Class 9 - Science - Gravitation Page 144
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Here, u = 49 m/s
v = 0 (at the highest point)
a = - 9.8 m/s²
◆ ¡) using 2aS = v² - u², we get
S = v² - u²/2a
= 0 - 49 × 49/ -2 × 9.8
= 122.5 m
Therefore, Maximum height = 122.5 m.
◆ ¡¡) Using, v = u + at, we get ,
t = v - u/ a
= 0 - 49/- 9.8
= 5 s.
ie; Time taken by the ball to reach the maximum height = 5 s.
The ball takes the same time to reach the ground from maximum height as time of ascent is equal to time of descent.
Therefore, Total time taken = 5 s + 5 s = 10 s.
v = 0 (at the highest point)
a = - 9.8 m/s²
◆ ¡) using 2aS = v² - u², we get
S = v² - u²/2a
= 0 - 49 × 49/ -2 × 9.8
= 122.5 m
Therefore, Maximum height = 122.5 m.
◆ ¡¡) Using, v = u + at, we get ,
t = v - u/ a
= 0 - 49/- 9.8
= 5 s.
ie; Time taken by the ball to reach the maximum height = 5 s.
The ball takes the same time to reach the ground from maximum height as time of ascent is equal to time of descent.
Therefore, Total time taken = 5 s + 5 s = 10 s.
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