A ball is launched with an initial velocity v_i = 25.0 m/s and a 50 degree angle

Question

A ball is launched with an initial velocity v_i = 25.0 m/s and a 50 degree angle from the top of the cliff at point. It takes t = 3.91 seconds for the ball to reach point 3. The coordinate system is given in this problem. (a) What is the numerical values of a_x and a_y for the ball in motion? State your answers without using any calculation in this part. (b) Find the horizontal travelling distance from point 1 to point 3. (c) Find horizontal traveling (at point 2) of the ball with respect to the origin shown in the graph. (d) What is the velocity in magnitude/direction form) when the ball reaches point 3? Explain without using any calculation in this part.

Explanation / Answer

(a) ax = 0, ay = -g = -9.81 m/s2

(b) Horizontal distance at point 3 is, d = (25.0 * cos50o) * 3.91 = 62.8 m

(c) Maximum height = h + (vi2sin250o/2g) = 50 + [25.02 * cos250o / (2 * 9.81)] = 63.2 m

(d) The horizontal velocity remains constant throughout the flight,

At point 3, the magnitude of the vertical velocity is same as that of the starting vertical velocity but its direction reverses i.e. directed downwards.

So, the velocity at point will be 25.0 m/s at an angle 50o below the x-axis.

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