Question

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (1) The maximum height to which it rises. (2) The total time it takes to return to the surface of the earth.

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Answer 1

$\bold{\underline{GIVEN :}}$

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

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

•g = 9.8m/s²

$\bold{\underline{SOLUTION :}}$

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

${\boxed{\red{\tt{H = 122.5 m}}}}$

Now consider a formula,

v = u + g × t

0 = 49 + (- 9.8) × t

– 49 = – 9.8t

${\boxed{\red{\tt{t = 5 sec}}}}$

(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.

Answer 2

Answer:

• 122.5m

• 10 sec

Explanation:

Initial velocity , u = 49 m/s

Final velocity , v = 0m/s

Maximum height to which it rises acceleration due to gravity = 9.8m/s²

An object is moving upwards , a = -9

We know that ,

v² - u² = 2 as

v² - u² = 2gh

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h = \dfrac{ {v}^{2} - {u}^{2} }{ - 2g}$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{ {49}^{2} - {0}^{2} }{2 \times 9.8}$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{2401}{19.6}$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:112.5m$

ii) Total time it takes to return to the surface of the earth.

from B to A:-

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:s = ut \dfrac{ - 1}{2} {gt}^{2}$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 122.5 = 49t - \dfrac{ - 1}{2} \times {9.8t}^{2}$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 49t \bigg(1 - \dfrac{t}{10} \bigg)$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:49t \bigg( \dfrac{10 - t}{10} \bigg)$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:1225 = 49t(10 - t)$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{1125}{49} = t(10 - t)$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:25 = 10t - {t}^{2}$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - {t}^{2} + 10t - 25 = 0$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {t}^{2} - 10t + 25 = 0$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: t = 5sec$

From A to B

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:s = ut + \frac{1}{2} {gt}^{2}$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:122.5 = (0)t + \frac{1}{2} \times 9.8 {t}^{2}$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:t = \dfrac{1225.5}{4.9} = 25$

$\sf \: \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:t = \sqrt{25} = \pm \: 5$

t = 5 sec

Total time = 5 + 5 = 10 sec