Question

a ball is thrown upward with velocity v from a height h above the ground. the time taken by the bal to hit the ground
a)root 2h/g
b)root 8h/g
c)(root v^2 +2gh)/g+v/g
d)v/g+root (2h/g)​

Answer 1

SOLUTION

Option (C) is correct

Given

• Height from which the ball is thrown : h

• Initial Velocity of the ball : v

According to sign convention,

Acceleration due to gravity is - g m/s² here

From second kinematic equation,

$\sf \: h = vt - \dfrac{1}{2} {gt}^{2} \\ \\ \longrightarrow \: \sf \: {gt}^{2} - 2vt + 2h = 0$

The above equation is of the form ax² + bx + c

Applying the Quartic Formula,

$\longrightarrow \: \sf \: t = \dfrac{ - ( - 2v) \pm \: \sqrt{( - 2v) {}^{2} - 4g( - 2h)} }{2g}$

Since,

Time can't be negative.

We will only consider

$\longrightarrow \: \sf \: t = \dfrac{2v + \sqrt{4 {v}^{2} + 8gh}}{2g} \\ \\ \longrightarrow \: \sf \: t \: = \dfrac{2(v + \sqrt{ {v}^{2} + 2gh}) }{2g} \\ \\ \longrightarrow \sf \: t \: = \dfrac{v \: + \sqrt{ {v}^{2} + 2gh} }{g} \\ \\ \longrightarrow \boxed{\boxed{\sf t = \dfrac{v}{g}\big[ {1 + \sqrt{1 + \frac{2gh}{v^2}}} \big]}}$