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83. A racquet ball is struck in such a way that it leaves the

racquet with a speed of 4.87 m/s in the horizontal

direction. When the ball hits the court, it is a horizontal

distance of 1.95 m from the racquet. Find the height of

the racquet ball when it left the racquet.

Answer :

Answer:

0.7864m

Explanation:

Ignore air resistance (the ball's horizontal speed stays constant), the time it takes for the ball to travel 1.95 m horizontally at the rate of 4.87 m/s.

t = 1.95 / 4.88 = 0.4s

This is also the time it takes to fall vertically with gravitational acceleration of g = 9.81 m/s2. We can use the following equation of motion to estimate the vertical distance:

[tex]h = gt^2/2[/tex]

[tex]h = 9.81*0.4^2/2 = 0.7864 m[/tex]

The height of the racquet ball  is 0.8 m.

Since the ball hits the court at a horizontal distance of 1.95 m from the racquet, hence the range is 1.95 m.

Range = velocity * time

1.95 = 4.87 * t

t = 0.4 s

Hence it took 0.4 s for the ball to land. The height (h) is:

h = (1/2)gt²

but g = acceleration due to gravity = 10 m/s², hence:

h = (1/2)*10*0.4² = 0.8 m

Therefore the height of the racquet ball  is 0.8 m.

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