A batted baseball leaves the bat at an angle of 30.0 degrees about the horizontal and is caught by an outfilder 375 ft from home plate at the same height from which it left the bat. (a) what was the initial speed on the ball? (b) how high does the ball rise about the point where is struck the bat?

Question

A batted baseball leaves the bat at an angle of 30.0 degrees about the horizontal and is caught by an outfilder 375 ft from home plate at the same height from which it left the bat. (a) what was the initial speed on the ball? (b) how high does the ball rise about the point where is struck the bat?

Answer 1

break up the initial speed into vertical and horizontal components.

Using the horizontal component, and distance, you know how long it was in the air

375=Vicos30*time
time=375/cos30

Now, knowing time in air..
hfinal=hinitia+visin30*time-1/2 g t^2

or
vi*sin30=1/2 g t
put in for time what you got before, and then solve for vi.

Answer 2

so for time i got 2431 s

and i used vi*sin30=1/2 g t
vi * -.9880 = 1/2 (9.8) (2431) +.9880

= 11912.888 is this correct

Answer 3

Hardly. Two thousand seconds in the air?

timeinair=375ft/Vi*.866=433/Vi
Notice I am working in English units.
Vi*1/2=1/2 (33ft/sec^2)433/Vi
vi= sqrt (33*433) ft/sec

check my thinking, and work.

Answer 4

To solve this problem, we can apply the principles of projectile motion. Let's break down the problem into two parts: (a) determining the initial speed of the ball and (b) calculating the maximum height it reaches.

(a) To find the initial speed of the ball, we need to consider the horizontal and vertical components of velocity separately.

The horizontal component of velocity remains constant throughout the entire flight. We can use the formula for horizontal distance to solve for the initial speed. The formula is:

Horizontal distance = Initial horizontal velocity × Time

In this case, the horizontal distance is given as 375 ft, and the time of flight is the same for both horizontal and vertical components.

Horizontal distance = 375 ft

Next, we need to find the vertical component of velocity. We can use the following formula to calculate the time of flight:

Time of flight = Vertical displacement / Vertical component of velocity

Since the vertical displacement is zero (the ball is caught at the same height it left the bat), we know that the time of flight is the same as the total time of flight for the ball. So, we can use the vertical component of velocity and the launch angle to find the initial speed of the ball.

Vertical component of velocity = Initial speed × sin(angle)

Now we can solve for the initial speed:

375 ft = Initial speed × cos(30°) × time

And

0 ft = Initial speed × sin(30°) × time - (1/2) × 9.8 m/s^2 × time^2

Since time is the same for both equations, we can solve for it:

375 ft / (Initial speed × cos(30°)) = time

Using this value of time, we can substitute it into our second equation:

0 ft = (Initial speed × sin(30°)) × (375 ft / (Initial speed × cos(30°))) - (1/2) × 9.8 m/s^2 × (375 ft / (Initial speed × cos(30°)))^2

Solving this equation will give us the initial speed of the ball.

(b) To calculate the maximum height the ball reaches, we need to determine the vertical displacement. We can use the equation for vertical displacement:

Vertical displacement = (Vertical component of velocity)^2 / (2 × gravitational acceleration)

Now that we have the initial speed, we can use the formula for the vertical component of velocity to calculate it.

Finally, we can substitute the value of the vertical component of velocity into the equation for vertical displacement to find the maximum height.

Following these steps, you should be able to determine the answers to both (a) and (b) of the problem.