Question

183. A ball is projected at an angle of 45° to the vertical with a
velocity of 20m/sec. The time taken by the ball to reach
maximum height and the max height reached respectively are:
A. 2 seconds, 20 m
B. 1.414 seconds, 20 m
C. 3 seconds, 10 m
D. 0.707 seconds, 10 m
​

Answer 1

Given:

• Angle made with vertical = 45°; so, angle made with horizontal = 90°- 45° = 45°therefore, Angle of projection, θ =  45°

• Initial velocity of ball, u = 20 ms⁻¹

To find:

• Time taken by ball to reach maximum height, t =?

• Maximum height reached by ball, h =?

Formula required:

• Formula for time taken to reach the maximum height by projectile

$\purple{\bigstar}\boxed{\sf{t=\dfrac{u\;sin\;\theta}{g}}}$

• Formula to calculate maximum height reached by projectile

$\purple{\bigstar}\boxed{\sf{h=\dfrac{u^2\;sin^2\theta}{2\;g}}}$

[ Where t is time taken to reach maximum height, h is maximum height, u is initial velocity, g is acceleration due to gravity, θ is angle of projection ]

Solution:

Using formula for calculating the time taken to reach maximum height

$\implies\sf{t=\dfrac{u\;sin\;\theta}{g}}$

$\implies\sf{t=\dfrac{20\times\;sin\;45^{\circ}}{10}}$

$\implies\sf{t=2\times sin\;45^{\circ}}$

$\implies\sf{t=2\times \dfrac{1}{\sqrt{2}}}$

$\implies\underline{\underline{\purple{\sf{t=\sqrt{2}=1.414\;\;sec}}}}$

Using formula for calculating the maximum height reached by projectile

$\implies\sf{h=\dfrac{u^2\;sin^2\theta}{2\;g}}$

$\implies\sf{h=\dfrac{(20)^2\times sin^2\;45^{\circ}}{2\times 10}}$

$\implies\sf{h=\dfrac{400\times sin^2\;45^{\circ}}{20}}$

$\implies\sf{h=20\times sin^2\;45^{\circ}}$

$\implies\sf{h=20\times \left(\dfrac{1}{\sqrt{2}}\right)^2}$

$\implies\underline{\underline{\purple{\sf{h=20\times \dfrac{1}{2}=10\;\;m}}}}$

Therefore,

• Time taken by ball to reach maximum height is 1.414 seconds.
• and, Maximum height  reached by Ball is 10 metres.