Question

Two projectiles are thrown with the same initial speed, one at an angle θ with respect to the level ground and the other at angle 90° − θ. Both projectiles strike the ground at the same distance from the projection point. Are both projectiles in the air for the same length of time?

Answer 1

Answer:

Height is different for both .

The time in air are different for both.

Step-by-step explanation:

Given that initial velocity is same

Lets take initial velocity = u

One at  an angle θ.

other at angle 90° − θ.

Also given that their range are same

$R_1=R_2$

$R_1=\dfrac{u^2sin2\theta }{g}$

$R_2=\dfrac{u^2sin2(90-\theta)}{g}$

$R_2=\dfrac{u^2sin(180-2\theta)}{g}$

We know that

sin(180° − θ)=sin θ

So

$R_2=\dfrac{u^2sin2\theta }{g}$

Height in the air

$h_1=\dfrac{u^2sin^2\theta}{2g}$

$h_2=\dfrac{u^2sin^2(90-\theta)}{2g}$

We know that

sin(90° − θ)=cos θ

$h_2=\dfrac{u^2cos^2\theta}{2g}$          

From above we can say that height is different for both .

Time:

$T_1=\dfrac{2usin\theta}{g}$

$T_2=\dfrac{2usin(90-\theta)}{g}$

sin(90° − θ)=cos θ

$T_2=\dfrac{2ucos\theta}{g}$

The time in air are different for both.