Question

A boy stands at the top of the building, with height h above the ground and throws a ball. The ball lands 30m from the building at an angle of 60degrees with respect to the horizontal ground after 2.3s .
A. Find the height of the building.
B. Find the initial velocity at which the ball is thrown and its angle of projection.
C. What is the maximum height rwach by the ball?

Answer 1

A. If we are at the top of the building and we see the point where the ball landed at an angle 60 degrees above the horizontal, the height of the building is

$$H=30\text{tan}60^\circ=51.96\text{ m}.$$

B. Find the horizontal component of the initial velocity:

$$v_x=\frac{x_{hor}}{t}=\frac{30}{2.3}=13.04\text{ m/s}.$$

At the same time these 2.3 seconds is time required for the ball to go up, reach the maximum height above the building $h$ and fall to the ground:

$$2.3=t_{up}+t_{down}.$$

$$t_{up}=\frac{v_y}{g},$$

$$h=\frac{v_y^2}{2g},\space t_{down}=\sqrt{\frac{2(h+H)}{g}}=\sqrt{\frac{2\Big(\frac{v_y^2}{2g}+H\Big)}{g}},$$

substitute these times to the equation with 2.3 and solve for $v_y$:

$$2.3=\frac{v_y}{g}+\sqrt{\frac{2\Big(\frac{v_y^2}{2g}+H\Big)}{g}},\\ \space\\ v_y=-11.32\text{ m/s}.$$

The ball was thrown down with initial velocity

$$v=\sqrt{v_x^2+v_y^2}=17.27\text{ m/s}$$

at an angle

$$\theta=\text{arctan}\frac{v_y}{v_x}=-33.24^\circ$$

with regard to the horizontal (i.e. below the horizontal).

C. Thus, the maximum height was the height of the building, or 51.96 m. Indeed, imagine the ball was thrown absolutely horizontally. Then the time of falling would be more than 3 seconds. It means that the ball had a vertical component of initial velocity looking down.