Question

31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far out would the ball travel? (8 points)

Answer 1

A) t = 0.55 s

B) x = 24.8 m

Step-by-step explanation:

A) We can find the time at which the ball will be in the air using the following equation:

`y_(f) = y_(0) + v_(0y)t - (1)/(2)gt^(2)`

Where:

`y_(f)` is the final height= 0

`y_(0)` is the initial height= 1.5 m

`v_(0y)` is the component of the initial speed in the vertical direction = 0 m/s

t: is the time =?

g: is the gravity = 9.81 m/s²

`0 = 1.5 m - (1)/(2)9.81 m/s^(2)t^(2)`

By solving the above equation for t we have:

`t = \sqrt{(2*1.5 m)/(9.81 m/s^(2))} = 0.55 s`

Hence, the ball will stay 0.55 seconds in the air.

B) We can find the distance traveled by the ball as follows:

`x_(f) = x_(0) + v_(0x)t + (1)/(2)at^(2)`

Where:

a: is the acceleration in the horizontal direction = 0

`x_(f)` is the final position =?

`x_(0)` is the initial position = 0

`v_(0x)` is the component of the initial speed in the horizontal direction = 45 m/s

`x_(f) = x_(0) + v_(0x)t + (1)/(2)at^(2)`

`x_(f) = 0 + 45 m/s*0.55 s + 0 = 24.8 m`

Therefore, the ball will travel 24.8 meters.

I hope it helps you!