Question

A ball is thrown vertically upwards from the top of a tower with a velocity of 20m/s. the ball strikes the ground after 5 seconds of it's throwing.what is the height of the tower and the velocity of the ball while striking the earth?(g=9.8m/s²


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Answer 1

Step-by-step explanation:

Given :

• Initial velocity (u) = 20m/s
• Distance (s) = ?
• Time (t) = ?
• Gravitational acceleration (a) = 9.8 m/s²

To find :

• Height of the tower
• Velocity of ball while striking the Earth - final velocity (v)

Formulae being use :

${\bold{⟼ \: \: \: \: \: \: \: \: \: \: }}{\bold{\underline{\boxed{\rm{S = -ut \: + \: \frac{1}{2}at²}}}}}$

${\bold{⟼ \: \: \: \: \: \: \: \: \: \: }}{\bold{\underline{\boxed{\rm{v = u \: + \: at}}}}}$

Where,

S = distance covered

u = initial velocity

a = acceleration due to gravity

t = time

v = final velocity

Solution :

❏ Using the law of motion to calculate height of the tower

${\bold{\sf{: \: ⟹ \: \: \: \: \: S = -ut \: + \: \frac{1}{2}at²}}}$

Substituting the values of initial velocity, time and acceleration due to gravity in above mentioned equation

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: S = \: -20 \: × \: (5) \: + \: \frac{1}{2} \: × \: 9.8 \: × \: (5)²}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: S \: = \: -100 \: + \: 1 \: × \: 4.9 \: × \: 25}}}$

${\bold{: \: ⟹ \: \: \: \: \: \: }}{\bold{\boxed{\boxed{\rm{\orange{S = 22.5 \: m}}}}}}$

${\bold{: \: ⟹ \: \: \: \: \: \: }}{\bold{\rm{\pink{ Height \: of \: the \: tower \: is \: 22.5 \: m }}}}$

❏ Again we have to use law of motion to calculate final velocity

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: v \: = \: u \: + \: at}}}$

Substituting the values of initial velocity, time and acceleration due to gravity in above mentioned equation

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: v \: = \: -20 \: + \: 9.8 \: × \: 5}}}$

${\bold{: \: ⟹ \: \: \: \: \: \: }}{\bold{\boxed{\boxed{\rm{\orange{v = 29 \: m/s}}}}}}$

${\bold{: \: ⟹ \: \: \: \: \: \: }}{\bold{\rm{\pink{Velocity \: of \: ball \: while \: striking \: the \: earth \: is \: 29 m/s }}}}$

Additional Information :

${\bold{\rm{⟼ \: \: \: \: \: \: v \: = \: u \: + \: at}}}$

${\bold{\rm{⟼ \: \: \: \: \: \: s \: = \: ut \: + \: \frac{1}{2}at²}}}$

${\bold{\rm{⟼ \: \: \: \: \: \: v² \: = \: u² \: + \: 2as}}}$

${\bold{\rm{⟼ \: \: \: \: \: \: s({t}^{th}) \: = \: u² \: + \: \frac{1}{2} \: a(2t \: - \: 1)}}}$

Answer 2

By using laws of motion we get

$\bold{s=−ut+ \frac{1}{2} at {}^{2}}$

u=20 m/s

By substituting the time and initial velocity in above equation and acceleration due ot gravity as 9.8 $m/s^2$

$\small\bold{s=−20×(5)+21×9.8×(5) {}^{2} }$

s=22.5m

Again by using

v=u+at

The velocity of the body is calculated as

v=−20+9.8×5

v=28 m/sec

Then velocity of the ball while stinking is 29 m/sec