A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
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The catcher can catch the ball at a height of 0.96 m from the ground.
The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.
Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.
The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.
Therefore,
Calculate the time t by substituting 18.44 m for x and 40 m/s for u.
The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.
Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.
Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,
Substitute 9.8 m/s² for g and 0.461 s for t.
The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.
Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,
The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>