Chapter - 3 Gravitation

Q&A -Ask Doubts and Get Answers

Q
Gravitation

Question:

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.

Answer:

Given:

Initial velocity, u = 49 m/s

Final speed v at maximum height = 0

Acceleration due to gravity g = -9.8 m/s² (g is negative as the ball is thrown up).

(i) Using the third equation of motion,

v^2 - u^2 = 2as

By substituting all the values in the above equation, we get

(0)^2 - (49)^2 = 2 \times -9.8 \times s

- 2401 = -19.6 \times s

s = \frac{2401}{19.6}

s = 122.5\ m

(ii) Let t be the time taken by the ball to reach the height 122.5 m.

Using the first equation of motion.

v = u + at

We get,

0 = 49 - 9.8t

49 = 9.8t

t = \frac{49}{9.8}

t = 5\ s

Time of ascent = Time of descent

Therefore, total time taken by the ball to return = 5 + 5 = 10 s


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