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?

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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

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