Question

a ball is thrown vertically upwards at 25m/s.gravity causes the ball to decelerate at 10m/s².calculate the maximum height the ball will reach ​

Answer 1

Answer:

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Explanation:

Let the point of projection of the ball be taken as origin of the coordinate system and upward direction as positive and downward direction as negative. At the highest point h, the velocity of the ball would become zero. We can get h using the relation: v² - u² = 2 g h. In our case the initial velocity u is +25 m/s, v= velocity at highest point =0 m/s; and g= -10 m/s² (negative sign as g is directed downward). So we get;

0² - 25² = - 2 × 10 × h ; ==> h = 25²/20 = 625/20 = 31.25 m. The highest point is 31.25 m above the point of projection.

To obtain velocity at 25 m height above the ground we again use the same relation. Now h= 25 m, u = 25 m/s, and v= ?

v² - 25² = 2 × -10 × 25 = -500 ; >or v² = 625 - 500 = 125 or v = ±√125 = +5√5 m/s while going up and -5√5 m/s while coming down from the highest point.

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