Question

1) A projectile is fired horizontally from the top of a cliff as shown: 1.00 km The projectile hits the ground 4.00 s later at a distance of 1.00 km from the base of the cliff. a 5 pts) What is the height of the cliff? b - 5 pts) What was the initial speed of the projectile? c - 5 pts) What is the speed of the projectile just before it hits the ground?

Answer 1

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From figure it is clear that the initial velocity is parallel to horizontal Time of flight T = 4 s horizontal range, R = 1 times 10^3 m a) we know that initial velocity along y - axis is as so h = u_y t + 1/2 gt^2 t = T u_y = 0 h = 1/2 g T^2 = 1/2 (9.8) (4)^2 = 78.4 m b) we know that t = x/u_x = distance in x direction/speed u_x = R when t = T so u_x = R/T = 10^3/4 = 250 m/s \r\n c) using equation of motion let v be the velocity just before the projectile hits ground v^2 = u_x^2 + 2gy = u_x^2 + 2g (1/2 gt^2) v^2 = u_x^2 + g^2 t^2 u_x = 250 m/s t = T v = squareroot 250^2 + (9.8 times 4)^2 = 253.05 m/s