Question

projectile is fired with a velocity of 320 m/s at an angle 30° to the horizontal. Find i) the time to reach
greatest height, ii) the horizontal range and iii) with the same velocity what is the maximum possible
range?

Answer 1

Given,

Initial Velocity ($u$) = 320 m/s

Angle of projection ($\theta$) = 30°

To Find,

i) The time to reach greatest height

ii) The horizontal range

iii) The maximum possible range with the same velocity

Solution,

By projectile motion we mean that when a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth.

i) Now, the required formula for obtaining the time to reach greatest height is,

$t=\frac{usin\theta}{g}$

where, $g$ = acceleration due to gravity = $9.8 m/s^{2}$

Therefore,

$t= \frac{320*sin30^{o}}{9.8} \\t= 16.33 s$

Hence, the required time to reach greatest height is 16.33 s.

ii) Horizontal range,

$R = \frac{u^{2}sin2\theta}{g} \\R = \frac{320^{2}sin60^{o}}{9.8} \\R = 9049$

Therefore, horizontal range of the projectile is 9.049 km.

iii) From the ii) problem, we can see that if the initial velocity remains constant, R varies with the the value of $\theta$.

R would be maximum for the maximum value of $sin2\theta$ i.e.,

$sin2\theta=sin 90^{o}=1$

so,

$2\theta=90^{o}\\\theta=45^{o}$

Therefore,

$R_{max} = \frac{u^{2}}{g} \\R_{max} = \frac{320^{2}}{9,8} \\R_{max} = 10449$

Hence, maximum possible range with the same velocity is 10.449 km.