Answer:
Height of the first projectile = 49.98 m
Height of the second projectile = 72.52 m
Explanation:
From the given information;
Two projectiles are thrown from the same point with the velocity of49m/s
First is projected making an angle θ with the horizontal
and the second at an angle of 90 - θ.
Thus; for the first height to the horizontal; we have;
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex] ----- (1)
the second height in the vertical direction is :
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex] -----(2)
However; the second is found to rise 22.5 m higher than the first; so , we have :
[tex]\dfrac{v^2 cos^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
Let's recall that :
Cos²θ = 1 - Sin²θ
Replacing it into above equation; we have:
[tex]\dfrac{v^2}{2g} - \dfrac{v^2 sin^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]\dfrac{v^2}{2g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } \dfrac{v^2}{g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } - \dfrac {9.8 \times 22.5}{(49)^2} = sin^2 \theta[/tex]
[tex]\dfrac{1}{2 } - \dfrac {220.5}{2401} = sin^2 \theta[/tex]
[tex]sin^2 \theta= 0.408[/tex]
From (1);
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]H_1 = \dfrac{49^2 \times 0.408}{2*9.8}[/tex]
[tex]H_1 = \dfrac{979.608}{19.6}[/tex]
[tex]\mathbf{H_1 =49.98 \ m }[/tex]
Height of the first projectile = 49.98 m
Similarly;
From(2)
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex]
[tex]H_2 = \dfrac{v^2 (1-sin^2 \theta)}{2g}[/tex]
[tex]H_2 = \dfrac{49^2 (1-0.408 )}{2 \times 9.8}[/tex]
[tex]H_2 = \dfrac{2401 (0.592 )}{19.6}[/tex]
[tex]H_2 = \dfrac{1421.392}{19.6}[/tex]
[tex]\mathbf{H_2 = 72.52 \ m}[/tex]
Height of the second projectile = 72.52 m
