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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?

User Babca
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1 Answer

1 vote

Answer:

Height is different for both .

The time in air are different for both.

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=(u^2sin2\theta )/(g)


R_2=(u^2sin2(90-\theta))/(g)


R_2=(u^2sin(180-2\theta))/(g)

We know that

sin(180° − θ)=sin θ

So


R_2=(u^2sin2\theta )/(g)

Height in the air


h_1=(u^2sin^2\theta)/(2g)


h_2=(u^2sin^2(90-\theta))/(2g)

We know that

sin(90° − θ)=cos θ


h_2=(u^2cos^2\theta)/(2g)

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

Time:


T_1=(2usin\theta)/(g)


T_2=(2usin(90-\theta))/(g)

sin(90° − θ)=cos θ


T_2=(2ucos\theta)/(g)

The time in air are different for both.

User Swalih
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