I need help with physics homework.?

During the time that the ball is in the air, its vertical velocity decreases at the rate of 9.8 m/s each second. The horizontal velocity is constant. The first step is to determine the vertical and horizontal components of the ball’s initial velocity.

Vertical = v * sin 25, Horizontal = v * cos 25

Let’s use the following equation to determine the time.

d = v * t, d = 301.5 meters, v = v * cos 25

301.5 = v * cos 25 * t

t = 301.5 ÷ (v * cos 25)

Since the ball’s vertical velocity is 0 m/s when it reaches the maximum height, let’s use the following equation to determine the time.

vf = vi – a * t, vf = 0, vi = v * sin 25, a = 9.8

0 = v * sin 25 – 9.8 * t

t = v * sin 25 ÷ 9.8

Let’s set these two equations equal to each other.

301.5 ÷ (v * cos 25) = v * sin 25 ÷ 9.8

Multiply both sides by 9.8.

2954.7 ÷ (v * cos 25) = v * sin 25

Multiply both sides by v * cos 25.

2954.7 = v^2 * cos 25 * sin 25

2 * sin 25 * cos 25 = sin 50

cos 25 * sin 25 = sin 50/2

2954.7 = v^2 * sin 50/2

5909.4 = v^2 * sin 50

v^2 = 5909.4 ÷ sin 50

v = √(5909.4 ÷ sin 50) = 87.83036967 m/s

To check this answer, let’s determine the vertical and horizontal components of the ball’s initial velocity.

Vertical = 87.83036967 * sin 25, Horizontal = 87.83036967 * cos 25

Let’s determine the time.

t = 301.5 ÷ (87.83036967 * cos 25) = 3.78624302 seconds

t = 87.83036967 * sin 25 ÷ 9.8 = 3.787624302 seconds

This proves that the initial velocity is correct.

U would not be previewing a question like this *early* unless you had some background in physics because it is not elementary but takes several steps to derive the solution equation plus knowledge of elementary motion.

Go to the source page under "Range of Trajectory" for more:

R = Vo²(sin 2Θ/g)

301.5 = Vo²(0.7660/9.81) = 0.07809Vo²

Vo² = 3861.0

62.1 m/s ANS