A tennis ball is thrown vertically upward with an initial velocity of +3.0 m/s.

a. What will the ball's speed be when it returns to its stating point?

b. How long will the ball take to reach its starting point?

c. Find the maximum height reached by the ball.

PLEASE I NEED STEPS

Respuesta :

Answer:

ok so i am not too giod at this but i will try

Explanation:

fisrt let me give tou a explation of velocity speed of motion, action, or operation; rapidity; swiftness. physics a measure of the rate of motion of a body expressed as the rate of change of its position in a particular direction with time.

Assuming no air restance the speed when the ball comes back to the starting point will be again 3ms but directed DOWNWARDS; we can express this saying that it will equal to −3ms adding a minus to indicate the downward direction.

To find the time of flight we use:

vf=vi+at

Where:

a is the acceleration of gravity (downwards, −9.8ms2);

vi=+8ms

vf=−8ms

So, we get:

−8=8−9.8t

−16=−9.8t

t=169.8≈1.6s

i am not good at this i tryed

(a)  the speed of the ball when it returns to its stating point is 3 m/s

(b) the time taken for the ball to reach the starting point is 0.31 s

(c) the maximum height reached by the ball is 0.46 m

The given parameters;

initial velocity of the ball, u = 3 m/s

(a) The velocity of the ball decreases as it moves upwards and eventually becomes zero when the ball reaches maximum height. As the ball moves downwards, the speed increases and eventually becomes maximum before the ball hits the ground.

Thus, the speed of the ball when it returns to its stating point is 3 m/s

(b) the time taken for the ball to reach the starting point is calculated as;

[tex]v_f = v_0 + gt\\\\3 = 0 + 9.8t\\\\9.8t = 3\\\\t = \frac{3}{9.8} \\\\t = 0.31 \ s[/tex]

(c) the maximum height reached by the ball.

at maximum height, the final velocity is zero;

[tex]v_f^2 = v_0^2 -2gh\\\\0 = 3^2 - (2\times 9.8)h\\\\19.6h = 9\\\\h = \frac{9}{19.6} \\\\h = 0.46 \ m[/tex]

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