Question

3. 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 1

Answer:

Given Initial velocity of ball, u=49 m/s

Let the maximum height reached and time taken to reach that height be H and t respectively.

Assumption: g=9.8 m/s

2

holds true (maximum height reached is small compared to the radius of earth)

Velocity of the ball at maximum height is zero, v=0

v

2

−u

2

=2aH

0−(49)

2

=2×(−9.8)×H

⟹H=122.5 m

v=u+at

0=49−9.8t

⟹t=5 s

∴ Total time taken by ball to return to the surface, T=2t=10 s

Answer 2

Explanation:

Given parameters

Initial velocity of the ball (u) = 49m/s.

The velocity of the ball at maximum height (v) = 0.

g = 9.8m/s2

Let us considered the time taken 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

(1) The maximum height to ball rises = 122.5 m

(2) The total time ball takes to return to the surface of the earth = 5 + 5 = 10 sec.