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?

Question

anonymous · Answer

a. 2.05 s

 b. 2.96 s

 c. 2.82 s

 d. 1.80 s

 e. 16.2 s

anonymous · Answer

Remember these equations -- they are very useful for beer pong.

The equation of motion for this system is

$$x(t) = x_0 + v_o \cos(\theta)t$$
$$y(t) = y_0 +v_0 \sin(\theta)t - (1/2)g t^2$$

So the thought here is, "at what time will the ball hit the ground"?  If you think about it, the only force in this problem is gravity.  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 y equation alone.  If that doesn't make sense, drop a ball with your left hand while throwing a ball with your right from the same height.  They'll hit the ground at the same time...  Here's the important insight.  When the ball hits the ground, it's height, or y, is zero.  So... set the y equation equal to zero on the left, and solve for t.  You'll need to use the quadratic equation.  Plug in, and you're done!