A ball is thrown at a 60.0° angle above the horizontal across level ground. It is released from a height of 2.00 m above the ground with a speed of 16.0 m/s. How long does the ball remain in the air before striking the ground?
Im going to give you the formula, so that you can replace the data with the correspondent variables. The formula is : y(t)=y0+v0sin(θ)t−(1/2)gt^2 Remember that all objects accelerate uniformly near the surface of the Earth. Therefore, the time that the ball will hit the ground is determined by the equation i gave you. Hope this can help you