Question

A ball thrown vertically upwards returns to the thrower after 8s. Calculate
the Velocity with which it was thrown. Given g= 9.8m/s2

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Answer 1

Answer:

• 39.2 m/s is the required answer .

Explanation:

Given:-

• Final velocity ,v = 0m/s (as the ball reached its maximum height)
• Total time taken by ball to return the initial position = 8 second .So, Time taken by ball to reach maximum height = 8/2 = 4 second.
• Acceleration due to gravity ,g = 9.8m/s²

To Find :-

• Initial velocity ,u

Solution:-

As we have to calculate the initial velocity of ball . So we can apply the first equation of motion .

• v = u +at

where,

• v is the final velocity
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

→ 0 = u + (-9.8) × 4

→ -u = -39.2

→ u = 39.2 m/s

• Hence, the initial velocity of the ball which it was thrown 39.2 m/s

Answer 2

Given : Final Velocity of ball will be 0 m/s [ As , the ball reach it's maximum height ] , Total time take by the ball to reach it's maximum height and returns to it's initial position is 8 seconds & Acceleration due to gravity is : g= 9.8m/s²

Exigency To Find : The Initial velocity [u] of ball .

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⠀⠀⠀⠀⠀❒ Finding initial velocity of ball :

Given that ,

• Final Velocity [v] of ball will be 0 m/s [ As , the ball reach it's maximum height ]

• Total time take by the ball to reach it's maximum height and returns to it's initial position is 8 seconds .

Therefore,

• Time taken by the ball to reach it's maximum height :

$\qquad :\implies \sf Time \:Taken \:[t]\:= \:\dfrac{8}{2}$

$\qquad :\implies \sf Time \:Taken\:[ t ] \:= \bf 4 \:\:\sf seconds$

$\dag\:\:\sf{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ First \:Equation _{(Motion)} \:: v = u + at }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , u is the Initial Velocity, v is the final velocity , a is the Acceleration , t is the time taken & a is the Acceleration.

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf v = u + at$

$\qquad:\implies \sf 0 = u + (- 9.8) \times 4$

$\qquad:\implies \sf 0 = u - 9.8 \times 4$

$\qquad:\implies \sf 0 = u - 39.2$

$\qquad:\implies \sf - u = - 39.2$

$\qquad:\implies \sf \cancel{-} u = \cancel{-} 39.2$ [ "-ve " sign will be eliminated from both side ]

$\qquad:\implies \sf u = 39.2$

$\qquad :\implies \frak{\underline{\purple{\:u = 39.2 \:m/s \:\:}} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Initial \:Velocity \:of\:Ball\:is\:\bf{39.2 \:m/s.}}}}$

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