A ball is thrown vertically upward with a speed of 24.0m/s. (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after it reaches its highest point? (d) What is its velocity when it returns to the level from which it started?

Expert Answers

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Start with finding the time it takes it to reach the highest point at which the velocity = 0 and it will start reversing its path due to gravity.
This can be found from the equation v = u - gt (note gravity works against the ball hence the -ve sign). v is final velocity which is 0, g is acceleration of gravity 9.81 m/s^2 , u is initital velocity of 24 m/s and t is the time it takes to attain a zero velocity
Therefore,
0 = 24 - 9.81 *t
solving, t = 24/9.81 = 2.45 sec

(a) The vertical distance it travels is how high it will go, which is represented by
s = ut - 1/2 *g*t^2 (note the -ve sign for g here as well)
= 24 * 2.45 - 1/2* 9.81 * 2.45^2 = 29.35m

(b) Time t = 2.45 sec al already calculated

(c) This can be obtained from same equation s = ut + 1/2 *g*t^2, except in this case g is +ve as gravity works for the fall and initial...

(The entire section contains 2 answers and 434 words.)

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