Question

A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? (8 = 9.8 m/s² )
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Answer 1

Given:

• Initial speed, u = 15 m/s

• Final speed, v = 0

• Acceleration due to gravity, g = - 9.8 m/s² ( retardation )

To be calculated:

Calculate the maximum height before it begins to fall ?

Formula used:

V² = u² + 2gh

Solution:

We know that for a free falling body :

v² = u² + 2gh

★ Substituting the values in the above formula,

we get :

( 0 )² = ( 15 )² + 2 ( - 9.8 ) × h

⇒ 0 = 225 - 19.6 × h

⇒ 0 = 225 - 19.6h

⇒ 19.6 h = 225

⇒ h = 225/19.6

⇒ h = 2250/196

⇒ h = 11.4 m

Thus, the ball will go to maximum height of 11.4 m.

Answer 2

$\huge{\underline{\pink{\tt{Given,}}}}$

• A Ball is thrown up with a speed (Initial Velocity) = 15 m/s
• Speed at the top most point (Final Velocity) = 0
• Acceleration Due to Gravity = -9.8 m/s²

(Here , the Ball is going up against the gravity so the value of g will be negative )

$\huge{\underline{\pink{\tt{To \ Find,}}}}$

• The Ball Maximum Height before of begin to fall .

$\huge{\underline{\pink{\tt{Solution :}}}}$

$\longrightarrow$ We know that Which formula Use for Free Falling Body :

$\bigstar \boxed{\sf{\red{v^{2} = u^{2} + 2gh}}}$

Now Put the Value in this Formula :

$\implies \sf{(0) = (15)^{2} + 2 \times (-9.8) \times h}$

$\implies \sf{0 = 225 - 19.6 \times h }$

$\implies \sf{0 = 225 - 19.6h}$

$\implies \sf{19.6h = 225}$

$\implies \sf{h = \dfrac{225}{19.6}}$

$\implies \boxed{\sf{h = 11.4 m}}$

Therefore,

$\boxed{\purple{\tt{The\:Ball\:will\:go\:the\:Maximum\:height\:of\:11.4 m}}}$

$\rule{200}2$