How to find vertical distance given speed and time?

[shlh]

A baseball is thrown horizontally off a cliff with a speed of 10 ms-1. What is the vertical distance, to the nearest tenth of a meter, that the ball has fallen after 4.7 seconds?

 

[mikeph]

The mathematics of this question are exactly the same as those of your other question, which I answered here: https://www.physicsforums.com/showthread.php?t=637116.

 

[shlh]

yes,but this ques are asking for VERTICAL instead of horizontal? is there any difference in calculation?

 

[mikeph]

It was vertical in the other question as well. The arrow's horizontal motion was irrelevant, all you were asked to do was find the vertical displacement during a known period of constant acceleration.

Although if it did happen to be horizontal, the same equation would still apply.

 

[PJC01]

To find the vertical distance, we can use the formula d = 1/2gt^2, where d is the vertical distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. However, in this scenario, the ball is thrown horizontally, so we need to use the horizontal velocity of the ball (10 m/s) to find the time it takes for the ball to reach the ground. This can be done using the formula t = d/v, where v is the horizontal velocity. Plugging in the values, we get t = d/10.

Now, we can substitute the value of t into the first formula to get d = 1/2 * 9.8 * (d/10)^2. Solving for d, we get d = 1/2 * 9.8 * (4.7/10)^2 = 1.08 meters. Therefore, the vertical distance the ball has fallen after 4.7 seconds is approximately 1.1 meters.

It is important to note that this calculation assumes no air resistance and a flat surface. In reality, the ball may experience air resistance and may be thrown from a cliff with an inclined surface, which could affect the vertical distance. Additionally, the accuracy of the result may also be affected by measurement errors and uncertainties.