Question

2 projectiles are fired simultaneously from ground level with same initial speed (u).Both cover same horizontal distance of 160m on reaching the ground level .One of them reaches 6 sec prior to the other.Only gravitational acceleration g=10m/s squared governs the motion of both the projectile.Calculate u.[/B]

Answer 1

Answer:

vx = 160 / t         Vx = 160 / T    where T = t + 6

160  = V^2 sin (2 theta1) / g = V^2 sin (2 theta2) / g   range formula

sin (2 theta1) = sin (2 theta2)

So (180 - 2 theta1) = 2 theta2     if the sines are to be equal

vx = 160 / t       Vx = 160 / T

vy = g t / 2    Vy = g T / 2    using time to fall from maximum height

Vx^2 + Vy^2 = vx^2 + vy^2     both have same initial speed

(160 / T)^2 +  (g T / 2)^2 = (160 t)^2 + (g t / 2)^2

160^2 (T^2 - t^2) = g^2 / 4 (T^4 t^2 - t^4 T^2)      collecting terms

= g^2 / 4 (T^2 t -  t^2 T) ( T^2 t + t^2 T) = g^2 T^2 t^2 / 4 (T - t) (T + t)

4 * 160^2 = g^2 T^2 t^2     dividing both sides by (T^2 - t^2)

2 * 160 = g t T = g t (t + 6) = g t^2 + 6 t

t^2 + 6 t - 32 = 0      divide by g = 10

t = 3.40    solving quadratic    so T = 9.40

vx = 160 / 3.4 = 47         Vx = 160 / 9.4) = 17

vy = 10 * 3.4 / 2 = 17       Vy = 10 * 9.4 / 2 = 47

tan theta1 = 47 / 17      theta 1 = 70.1

tan theta2 = 17 / 47     theta2 = 19.9

check:   180 - (2 * 70.1) = 39.8    which is 2 * theta2

sin 140.2 = .64 = sin 39.8

Using 160 = V^2 sin 39.8 / 10

V^2 = 1600 / .64           or V = 50 m / s

50 * cos 19.8 = 47      50 * cos 70.1 = 17       etc.