Question

a ball is thrown vertically upwards with speed 35/ms calculate the maximum height attained by the ball take g = 10ms​

Answer 1

Explanation:

u= 35m/s.

g = -10 m/s^2

v= 0 m/s

as,

v^2 = u^2 - 2gh

0 = 35*35 - 2*10h

h = 35×35/20

= 35× 7/4

= 61.25m

Answer 2

GIVEN:-

• u = 35 m/s

TO FIND:-

• Maximum height = ?

By Using 3rd Kinematical equation :-

$\\ \leadsto \underline{ \boxed{ \sf {v}^{2} - {u}^{2} = 2as}} \bigstar$

After attaining the maximum height, the ball comes to rest momentarily which means the final velocity of the ball becomes 0 after covering the maximum height.

The ball is thrown upwards i.e opposite to the direction of Accelaration due to gravity. Hence, we'll assume the Accelaration to be negative.

$\\ \quad \dag \underline{ \sf substitute \: the \: values \: in \: the \: formula : - }$

↝ v² - u² = 2as

↝ (0)² - (35)² = 2 × (- 10) × S

↝ -1225 = -20 × S

↝ 1225 = 20 × S

↝ S = 1225/20

$\\ \therefore\boxed{ \sf s = 61.25 \: m} \bigstar$

hence, the maximum height is 61.25 m

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