Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. A shot put is thrown upward with a velocity of 35 ft./sec. at a height of 4 ft. and an angle of 40°. How long will it take for the shot put to be a horizontal distance of 40 ft. from the person throwing it? Acceleration due to gravity is 32 ft./s2 . Round your answer to the nearest hundredth.

Answer 1

To solve this problem, we need to break down the motion of the shot put into horizontal and vertical components.

First, let's calculate the initial vertical velocity (Vy) and horizontal velocity (Vx):
Vy = 35 ft/sec * sin(40°) = 22.52 ft/sec
Vx = 35 ft/sec * cos(40°) = 26.92 ft/sec

Next, let's find the time it takes for the shot put to reach a vertical height of 0 ft (when it reaches the ground). We can use the formula for vertical motion:
y = yo + Voy*t - 0.5*g*t^2
where y = 0, yo = 4 ft, Voy = 22.52 ft/sec, g = 32 ft/s^2

0 = 4 + 22.52*t - 0.5*32*t^2
0 = 4 + 22.52*t - 16*t^2
16*t^2 - 22.52*t - 4 = 0

Using the quadratic formula, we find:
t = (22.52 ± sqrt((22.52)^2 - 4*16*(-4))) / (2*16)
t = (22.52 ± sqrt(506.86 + 256)) / 32
t = (22.52 ± sqrt(762.86)) / 32
t ≈ (22.52 ± 27.61) / 32

We take the positive solution since time cannot be negative:
t ≈ (22.52 + 27.61) / 32
t ≈ 1.67 sec

Now, we can find how far the shot put travels horizontally in that time:
x = Vx * t
x = 26.92 ft/sec * 1.67 sec
x = 44.99 ft

Since the shot put travels more than 40 ft horizontally in 1.67 sec, it will take less than 1.67 sec to travel 40 ft horizontally.

By using linear approximation, we can say that the shot put will travel 40 ft horizontally when it's in the air for about 1.55 seconds.

Answer 2

My answer was 1.49

Answer 3

I apologize for the oversight. Let's go through the calculation again to find the time it takes for the shot put to reach a horizontal distance of 40 ft:

Using the horizontal motion formula:
x = Vx * t
40 ft = 26.92 ft/sec * t
t = 40 ft / 26.92 ft/sec
t ≈ 1.49 sec

Therefore, it will take approximately 1.49 seconds for the shot put to reach a horizontal distance of 40 ft from the person throwing it. Thank you for pointing out the error.