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
[tex]R_1=R_2[/tex]
[tex]R_1=\dfrac{u^2sin2\theta }{g}[/tex]
[tex]R_2=\dfrac{u^2sin2(90-\theta)}{g}[/tex]
[tex]R_2=\dfrac{u^2sin(180-2\theta)}{g}[/tex]
We know that
sin(180° − θ)=sin θ
So
[tex]R_2=\dfrac{u^2sin2\theta }{g}[/tex]
Height in the air
[tex]h_1=\dfrac{u^2sin^2\theta}{2g}[/tex]
[tex]h_2=\dfrac{u^2sin^2(90-\theta)}{2g}[/tex]
We know that
sin(90° − θ)=cos θ
[tex]h_2=\dfrac{u^2cos^2\theta}{2g}[/tex]
From above we can say that height is different for both .
Time:
[tex]T_1=\dfrac{2usin\theta}{g}[/tex]
[tex]T_2=\dfrac{2usin(90-\theta)}{g}[/tex]
sin(90° − θ)=cos θ
[tex]T_2=\dfrac{2ucos\theta}{g}[/tex]
The time in air are different for both.
