Calculating the Distance Dropped by a Baseball Thrown at 116 km/h

[shenwei1988]

a) A pitcher throws a baseball horizontally at 116 km/h toward home plate, which is 18.4 m away. How far will the ball drop due to gravity by the time it reaches home plate?

g=9.81m/s^2
v=116km/h


i tried to solve this question, but i don't know where to get the high.
116km/h=32.22m/s
18.4/32.22=0.57s
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please help me, thank you! i am so confuse with the question it actually ask.

 

[MaZnFLiP]

Basically, the question that it is asking is that it wants you to find how far the ball dropped since it left the pitcher's hand. What I did was that i wrote out all my horizontal givens and my vertical givens and I just solved for [tex]\Delta[/tex]y after I got all the other information.

 

[Moetasim]

Hello,

I understand your confusion with this question. Let me explain how to calculate the distance dropped by the baseball due to gravity using the given information.

First, we need to convert the velocity of the baseball from kilometers per hour (km/h) to meters per second (m/s). This can be done by multiplying the velocity by 1000/3600, which gives us 32.22 m/s.

Next, we can use the formula d= ½ * g * t^2, where d is the distance dropped, g is the acceleration due to gravity (9.81 m/s^2), and t is the time taken for the ball to travel the given distance.

We already know the time, which is 0.57 seconds (calculated by dividing the distance of 18.4 meters by the velocity of 32.22 m/s).

So, plugging in the values in the formula, we get d= ½ * 9.81 * (0.57)^2 = 1.61 meters.

Therefore, the ball will drop 1.61 meters due to gravity by the time it reaches home plate.

I hope this helps to clarify the question and the process of calculating the distance dropped. Keep up the good work with your scientific thinking!