Find the height from which the ball was thrown.

Question

A ball is tossed from an upper story window of a building. The ball is given an initial velocity of 6.6 m/s at an angle of 23⁰ below the horizontal. It strikes the ground 3.00 s later. Find the height from which the ball was thrown.

Answer 1

We can use our kinematic equations of motion to solve this problem. First, we need to identify the appropriate equation. We are trying to solve for the height from which a ball was thrown, and we are given an initial velocity and time it takes for the ball to hit the ground. The equation which contains these variables is:

Δy = v_0y*t + 1/2*a*t²

Some of these variables we know simply from reading the problem statement.

t = 3 seconds

a = -9.81 m/s² (acceleration due to gravity)

It is important to note that we are given an initial velocity of 6.6 m/s, but this is the total magnitude at an angle 23 degrees below the horizontal. We need to use trigonometry to determine the horizontal, and more importantly, vertical components of this velocity.

v_0x = 6.6*cos(23 deg) = 6.08 m/s (we do not explicitly need this value to solve the problem, but it is good practice to calculate, also note this horiz component of velocity remains constant throughout the ball's flight)

v_0y = 6.6*sin(23 deg) = 2.58 m/s, it is very important to note that 23 degrees below the horizontal points "downward", so this velocity should actually be -2.58 m/s.

Now, we simply substitute our variables into our equation to solve for the change in height (height from which ball was thrown).

Δy = (-2.58 m/s)*(3 s) + 1/2*(-9.81 m/s²)*(3 s)²

Δy = -51.9 m (negative sign is due to y_f - y_i = 0 m - 51.9 m)