A man throws a hall upwards, what is the maximum height h obtained by the ball?

Question

A man throws a hall upwards, what is the maximum height h obtained by the ball? Comments An illustration of the contrived system is provided in Fig. 1. This is a one-dimensional problem-the ball can move upward or downward. We can represent the "up" direction in terms of the "up" arrow as shown in Fig. 1. We can represent the downward direction in a similar fashion. We call the "up" arrow +e_x and the down arrow -e_x. The ball has a mass m equal to m = 145 g, the typical mass of a baseball. The ball is pitched upward and released at a height of x_0 = 3m above the ground with an initial speed of v_0. In general, the velocity v(t) has a magnitude |v(t)| and a direction that can both vary with time. The number |v(t)| is called the speed. We use the shorthand notation v_0 for the initial speed. We could just as well use the symbol |v(t = 0)| to represent the initial speed-both v_0 and |v(t = 0)| have the same meaning-namely the value of the ball's speed at that particular moment in time when we start the imaginary clock. In our case, the initial speed is equal to v_0 = 30 m/s We will neglect the effects of air resistance and consider the gravitational influence alone. In this case the velocity has the form FIG. 1: Sketch of the system described in Problem 1. A ball of mass m is thrown perfectly upward; the subsequent motion is governed by the ball's inertia and the influence of gravity. You are asked to determine the height h of greatest ascent.

Explanation / Answer

At the greatest height velocity become zero.

Using relation v2 - vo2 = 2 (-g) h

02 - 302 = 2 (-9.8) h

h = 45.92 m

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