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

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

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

$y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}$    

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

$0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}$

By solving the above equation for t we have:

$t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s$  

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

                             

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

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

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                                                                                            

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

$x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m$

Therefore, the ball will travel 24.8 meters.

I hope it helps you!