Question

63. 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:

v = 49m/s

Explanation:

Hmax = u^2/2g

= 49 × 49 / 2 ×9.8

= 122.5

time of flight = 2usin⊙/g

= 2×49×1/ 9.8 ( ⊙=90 as height maximum)

= 10

Answer 2

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

$\bf{\underline {Given:}}$

Initial velocity of the ball, $u\:=\:49 \:m\: {s}^{-1}$

Final velocity of the ball,$v = 0$

Acceleration due to gravity, $g = -9.8 m {s}^{-2}$

(i) Let $h$ be the maximum height attained by the ball.

According to the equation of motion under gravity,

${v}^{2} -{v}^{2} =2 \:gh$

Substituting these values, we get

$0-{(49)} ^{2} = 2 x (-9.8) x h$

$h =\dfrac{49×49}{2x9.8}=\bf {122.5\: m}$

(ii) Height achieved by the ball,$h= 122.5 m.$

Now, let 'f be the time taken by the ball to reach the height 122.5 m, then

According to the equation of motion,

$v = u + gt$

Substituting the values, we get

$0 = 49 + t ×(-9.8)$

$9.8t=49$

$t = \dfrac{49}{9.8}=\bf{5\:s}$

But, time of ascent = Time of descent

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