Question

Two projectiles are thrown from the same point with the velocity of49ms-1. First is

projected making an angle with the horizontal and the second at an angle of (90- ). The

second is found to rise 22.5m higher than the first. Find the heights to which each will rise?

Answer 1

Height of the first projectile = 49.98 m

Height of the second projectile = 72.52 m

Step-by-step 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;

`H_1 = (v^2 sin^2 \theta)/(2g)` ----- (1)

the second height in the vertical direction is :

`H_2 = (v^2 cos^2 \theta)/(2g)` -----(2)

However; the second is found to rise 22.5 m higher than the first; so , we have :

`(v^2 cos^2 \theta)/(2g)= 22.5 + (v^2 sin^2 \theta)/(2g)`

Let's recall that :

Cos²θ = 1 - Sin²θ

Replacing it into above equation; we have:

`(v^2)/(2g) - (v^2 sin^2 \theta)/(2g)= 22.5 + (v^2 sin^2 \theta)/(2g)`

`(v^2)/(2g) - 22.5 = (v^2 sin^2 \theta)/(g)`

`(1)/(2 ) (v^2)/(g) - 22.5 = (v^2 sin^2 \theta)/(g)`

`(1)/(2 ) - \frac {9.8 * 22.5}{(49)^2} = sin^2 \theta`

`(1)/(2 ) - \frac {220.5}{2401} = sin^2 \theta`

`sin^2 \theta= 0.408`

From (1);

`H_1 = (v^2 sin^2 \theta)/(2g)`

`H_1 = (49^2 * 0.408)/(2*9.8)`

`H_1 = (979.608)/(19.6)`

`\mathbf{H_1 =49.98 \ m }`

Height of the first projectile = 49.98 m

Similarly;

From(2)

`H_2 = (v^2 cos^2 \theta)/(2g)`

`H_2 = (v^2 (1-sin^2 \theta))/(2g)`

`H_2 = (49^2 (1-0.408 ))/(2 * 9.8)`

`H_2 = (2401 (0.592 ))/(19.6)`

`H_2 = (1421.392)/(19.6)`

`\mathbf{H_2 = 72.52 \ m}`

Height of the second projectile = 72.52 m