Question

A ball is projected with a speed of 20m/s at an angle with the horizontal in order to have maximum range the velocity at highest point must be

Answer 1

When the velocity becomes perpendicular to initial velocity following condition must hold true.

gt sin30 = v ( think why ? Hint: geometric method of vector addition )

t = v/ g× sin30

= 20/9.8× 0.5

= 4.08 seconds

Answer 2

Given:

Speed of the ball $=20\frac { m }{ s }$

In order to have the maximum range, the angle should be a = 45°

Component of the velocity, along the x-axis at any instant is given by $V_{ x }=u\times cos a$

It means that the horizontal component of the velocity does not change throughout the projectile motion.

$V_{ x }=20\quad Cos45$

$V_{ x }=20(\frac { 1 }{ \sqrt { 2 } } )$

$V_{ x }=10\times \frac { 2 }{ \sqrt { 2 } } =10\sqrt { 2 } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =10{ \sqrt { 2 }seconds$.