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guapka [62]
4 years ago
6

Most earthquakes occur along or near the edges of the earth's

Physics
1 answer:
Pepsi [2]4 years ago
8 0
Most earthquakes occur along or near the edges of the earth's lithospheric<span> plate. </span>
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3 years ago
A tennis player tosses a tennis ball straight up and then catches it after 1.25 s at the same height as the point of release.
Alenkasestr [34]

Answer:

A. 9.8 m/s²

B. Zero

C. 6.125 m/s

D. 1.91 m

Explanation:

From the question given above, the following data were obtained:

Time (T) spent in the air = 1.25 s

A. Determination of the acceleration of the ball.

From the description given in question above, the motion of the tennis ball is motion under gravity. Hence, the ball will experience an acceleration due to gravity of 9.8 m/s²

B. Determination of the velocity at maximum height.

Maximum height is the greatest point reached by the tennis ball above the ground. At maximum height, the velocity of the tennis ball is zero since it has no further force to propel it upward.

C. Determination of the initial velocity of the ball.

We'll begin by calculating the time taken to reach the maximum height. This can be obtained as follow:

Time (T) spent in the air = 1.25 s

Time (t) to reach the maximum height =?

T = 2t

1.25 = 2t

Divide both side by 2

t = 1.25 / 2

t = 0.625 s

Finally, we shall determine the initial velocity of the ball. This can be obtained as follow:

Time (t) to reach the maximum height = 0.625 s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 (at maximum height)

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 0.625)

0 = u – 6.125

Collect like terms

0 + 6.125 = u

u = 6.125 m/s

Thus, the initial velocity of the ball is 6.125 m/s

D. Determination of the maximum height.

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 (at maximum height)

Initial velocity (u) = 6.125 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 6.125² – (2 × 9.8 × h)

0 = 37.52 – 19.6h

Collect like terms

0 – 37.52 = – 19.6h

– 37.52 = – 19.6h

Divide both side by – 19.6

h = – 37.52 / – 19.6

h = 1.91 m

Thus, the maximum height reached by the ball is 1.91 m

3 0
3 years ago
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