Question

A boy standing on top of a building throws a small ball from a height of H₁ = 45.0 m. (See figure.) The ball leaves with a speed of 26.9 m/s, at an angle of 61.0 degrees from the horizontal, and lands on a building with a height of H₂ = 12.0 m. Calculate for how long the ball is in the air. (Neglect air friction, and use g = 9.81 m/s².)

Answer 1

The ball follows the projectile motion equations. In this kind of motion, you can work with the x and y coordinates separately. The equation for the position along the y axis is:

where y is the height (vertical coordinate of the position) at time t, y_o is the initial height, v_o is the magnitude of the initial velocity, theta is the angle for this initial velocity, and g is the magnitude of the acceleration due to gravity.

You are looking for the time when the height is 12.0m , that would be the value for y, and you have to solve for t.

Replacing the values into the equation you have:

And this is a quadratic equation, that you have to set equal to zero to solve it. So simplifying the numbers:

Moving everything to the left side:

Now you can use the quadratic formula to find the solutions:

where:

So:

Then the two solutions are:

and

The quadratic equation yields two solutions, but only one of them is valid in this question, because we are starting the problem at t=0, so the negative solution makes no sense for this problem.

The answer is: