Question

question 1
you are launching water balloons at a rival team using a large slingshot. the other team is set up on the opposite side of a flat - topped building that is 30.0 ft tall and 50.0 ft wide. your reconnaissance team has reported that the opposition is set up 10.0 m from the wall of the building. your balloon launcher is calibrated for launch speeds that can reach as high as 115 mph at angles between 0 and 85.0° from the horizontal. since a direct shot is not possible (the opposing team is on the opposite side of the building), you plan to splash the other team by making a balloon explode on the ground near them.
if your launcher is located 55.0 m from the building (opposite side as the opposing team), what should your launch velocity be ((a) magnitude and (b) direction) to land a balloon 5.0 meters beyond the opposing team with maximum impact (i.e. maximum vertical speed)?

Explanation:

Step1: Convert units

First, convert the height of the building from feet to meters. 1 ft = 0.3048 m, so $h = 30.0\times0.3048=9.144$ m. The width of the building in meters is $w = 50.0\times0.3048 = 15.24$ m. Also, convert the speed from mph to m/s. 1 mph = 0.44704 m/s, so $v_{max}=115\times0.44704 = 51.40$ m/s.

Step2: Determine the horizontal - range

The horizontal distance from the launcher to the target location (5.0 m beyond the opposing team) is $x=55.0 + 15.24+10.0 + 5.0=85.24$ m.

Step3: Analyze the vertical - motion

For maximum vertical speed on impact, the balloon should be launched at the maximum allowable angle $\theta = 85.0^{\circ}$. The equations of motion for projectile - motion are: $x = v_0\cos\theta t$ and $y - y_0=v_0\sin\theta t-\frac{1}{2}gt^2$. Since $y - y_0=- 9.144$ m (negative because the target is below the launch - height), and $x = v_0\cos\theta t$, then $t=\frac{x}{v_0\cos\theta}$.
Substitute $t$ into the vertical - motion equation: $y - y_0=v_0\sin\theta\frac{x}{v_0\cos\theta}-\frac{1}{2}g(\frac{x}{v_0\cos\theta})^2$.
$y - y_0=x\tan\theta-\frac{gx^{2}}{2v_0^{2}\cos^{2}\theta}$.
We know $y - y_0=-9.144$ m, $x = 85.24$ m, and $\theta = 85.0^{\circ}$.
$\cos\theta=\cos85.0^{\circ}=0.0872$, $\sin\theta=\sin85.0^{\circ}=0.9962$, $\tan\theta=\tan85.0^{\circ}=11.43$.
$-9.144=85.24\times11.43-\frac{9.8\times(85.24)^{2}}{2v_0^{2}\times(0.0872)^{2}}$.
$85.24\times11.43 = 974.39$.
$-9.144=974.39-\frac{9.8\times7266.86}{2v_0^{2}\times0.0076}$.
$\frac{9.8\times7266.86}{2v_0^{2}\times0.0076}=974.39 + 9.144=983.534$.
$\frac{71215.23}{0.0152v_0^{2}}=983.534$.
$v_0^{2}=\frac{71215.23}{0.0152\times983.534}$.
$v_0^{2}=\frac{71215.23}{14.95}$.
$v_0^{2}=4763.69$.
$v_0 = 69.0$ m/s.

Answer:

(a) The magnitude of the launch velocity $v_0 = 69.0$ m/s.
(b) The direction of the launch velocity $\theta = 85.0^{\circ}$ above the horizontal.