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?
2 answers
The ball's horizontal and vertical components are independent of each other. The 11m per sec makes no difference to your calculation. You know the starting vertical vel( 0m per sec); the time,(4.8sec) and the acceletation (g). Use v=u+at to get the velocity at 4.8 sec, then find an equation with all those, and distance in. solve for distance.
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
1 answer