Question

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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Answer 1

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

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

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Answer 2

Answer:

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate The maximum height to which it rises. (2) The total time it takes to return to the surface of the earth.

here is the 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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