A baseball is thrown horizontally off a cliff with a speed of 11 ms^-1. What is the vertical distance, to the nearest tenth of a meter, that the ball has fallen after 4.8 seconds?
use the formula s(distance) =ut + 1/2 at^2 . u is the m^-1 . a is the acceleration aka your gravity, it will be negative since the object will slowly descend downwards and t is the time after it falls. I got time for quarantine so I'm just gonna help you there so fill in your value in the formula. Hope it helps you
To find the vertical distance that the baseball has fallen after 4.8 seconds, we need to calculate the height using the equation of motion. The key here is to realize that the initial vertical velocity is 0 (since the ball is thrown horizontally), and the only force acting on the ball is gravity.
We can use the following equation to find the height (h):
h = (1/2) * g * t^2
where:
h = height (vertical distance)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (4.8 seconds in this case)
Substituting the given values into the equation, we have:
h = (1/2) * 9.8 * (4.8)^2
= 0.5 * 9.8 * 23.04
= 112.896
Rounding the answer to the nearest tenth, the vertical distance that the ball has fallen after 4.8 seconds is approximately 112.9 meters.