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1 year ago

A tennis player tosses a tennis ball straight up and then catches it after 1.64 s at the same height as the point of release.

Physics

1 answer:

ipn [44]1 year ago

4 0

A. The acceleration of the ball while it is in flight?

magnitude is 0 m/s² (magnitude is zero)

B. The velocity of the ball when it reaches its maximum height is 0 m/s (magnitude is zero)

C. The initial velocity of the ball 8.036 m/s upward

D. The maximum height reached by the ball is 3.29 m

<h3>A. How to determine the acceleration in the flight</h3>

Considering that the ball came to rest after 1.64s, it means the entire acceleration of the flight is zero as the ball was not moving in any form again.

<h3>B. How to determine the velocity at maximum height</h3>

At maximum height, the velocity of the ball is zero as it no longer has magnitude to keep going upwards. Hence the ball begins to ball down.

<h3>C. How to determine the initial velocity</h3>

Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) = 0 m/s
Time of flight (T) = 1.64 s
Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
Initial velocity (u) =?

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

0 = u - (9.8 × 0.82)

0 = u - 8.036

Collect like terms

u = 0 + 8.036

u = 8.036 m/s upward

<h3>D. How to determine the maximum height reached by the ball</h3>

Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h)

h = ½gt²

h = ½ × 9.8 × 0.82²

h = 3.29 m

Learn more about motion under gravity:

brainly.com/question/20385439

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A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat

Tanzania [10]

If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 $\text{m/s}^{2}$ .

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is $\omega$ =5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, $a=\omega ^2 r$ .

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

$\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$