An attacker at the base of a castle wall 3.70 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.70 m high throws a rock straight up with a speed 8.30 m/s from a height of 1.68 m above the ground.

a)Will the rock reach the top of the wall?

yes

b) What is the speed at the top?If not what initial speech must it have to reach the top?

c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 8.30 m/s and moving between the same two points?

I got 6.618 for part b using the two equations x+xo+v0t+1/2at^2 and v+vo+at
This answer was incorrect. Any help would be appreciated. Thanks.

Explanation / Answer

(a) the speed of the ball V = 8.30 m/s the height reached the ball in this speed, h = v^2/ 2g = 3.514 m the total distance traveled by the ball H = 1.68+3.514 = 5.194 m hence the rock reach the wall, (yes) (b) the speed of the ball at top, V^2 = V0^2 - 2*9.8*(3.70-1.68) V^2 = 68.89 - 39.59 V1 = 5.41 m/s (c) the spped of the boll at bottem, V^2 = V0^2 + 2*9.8*(3.70-1.68)     V2 = 10.41 m/s the speed's difference = 10.41-5.41 = 5 m/s

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