Question

When a ball thrown vertically upward provide to a distance of 19.6 M find the velocity of ball and time taken by it and rise the height

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Initial\:velocity=14\sqrt{2}\:m/s}}$

${\bold{\therefore Time=1.4\sqrt{2}\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a ball thrown vertically upward provide to a distance of 19.6 m.

• We have to find the velocity of ball and time taken by it.

$\underline \bold{Given : } \\ \implies Final \: velocity(v) = 0 \:m/s\\ \\ \implies Height(s) = 19.6 \: m \\ \\ \implies Acceleration(a) = -g = - 10 \: m /{s}^{2} \\ \\ \underline \bold{To \: Find : } \\ \implies Initial \: velocity(u) = ? \\ \\ \implies Time \: taken(t) = ?$

• According to given question :

$\bold{By \: second \: equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {0}^{2} = {u}^{2} + 2 \times ( - 10) \times 19.6 \\ \\ \implies {u}^{2} = 392 \\ \\ \implies u = \sqrt{392} \\ \\ \bold{\implies u = 14 \sqrt{2} \: m/s} \\ \\ \bold{By \: first \: equation \: of \: motion : } \\ \implies v = u + at \\ \\ \implies 0 = 14 \sqrt{2} + ( - 10) \times t \\ \\ \implies \cancel- 14 \sqrt{2} = \cancel- 10t \\ \\ \implies t = \frac{14 \sqrt{2} }{10} \\ \\ \bold{ \implies t = 1.4 \sqrt{2} \: sec }$

Answer 2

$\boxed{\sf\green{Answer}}$

$\sf u =14\sqrt{ 2}m{s}^{-1}$

$\sf t = 2s$

$\boxed{\sf \green{Explanation}}$

Given:

• S = 19.6m
• a = -g
• v = 0

To Find:

Velocity (v) and time taken(t)

Solution:

$\red{\boxed{\blue{{v}^{2}- {u}^{2} = 2as}}}$

on putting the values

$\sf{0}^{2}- {u}^{2} = 2(-9.8 m{s}^{-2})(19.6m )$

$\sf {u}^{2} = 2(10 m{s}^{-2})(19.6m)$

$\sf {u}^{2} = 2(196{m}^{2}{s}^{-2})$

$\sf u =\sqrt{ 2(196{m}^{2}{s}^{-2})}$

$\sf u =14\sqrt{ 2}m{s}^{-1}$

Now,

$\green{\boxed{\pink{v = u+at}}}$

$\sf 0 =14\sqrt{ 2}m{s}^{-1} + (-10m{s}^{-2})t$

$\sf 14\sqrt{ 2}m{s}^{-1} = (10m{s}^{-2})t$

$\sf t = \dfrac{14\sqrt{ 2}m{s}^{-1}} {10m{s}^{-2}}$

$\sf t = 1.4\sqrt{ 2}s$

$\sf t = (\sqrt{ 2})(\sqrt{ 2})s$

$\sf t = 2s$