Question

Some enterprising physics students working
on a catapult decide to have a water balloon
fight in the school hallway. The ceiling is of
height 3.3 m, and the balloons are launched
at a velocity of 11 m/s.
The acceleration of gravity is 9.8 m/s².
At what angle must they be launched to
just graze the ceiling?
Answer in units of °.

Answer 1

Answer:

$47^\circ$

Explanation:

To find the launch angle at which the water balloons just graze the ceiling, we can use the following kinematic equation for projectile motion:

$\Large\boxed{\boxed{ h = \dfrac{{u^2 \sin^2 \theta}}{{2g}} }}$

where:

• $h$ is the height of the ceiling (3.3 m),
• $u$ is the initial velocity of the water balloon (11 m/s),
• $\theta$ is the launch angle,

$g$ is the acceleration due to gravity (9.8 m/s²).

We want the water balloon to just graze the ceiling, which means it reaches a maximum height equal to the height of the ceiling.

So, we can set $h$ equal to 3.3 m:

$3.3 = \dfrac{{(11)^2 \sin^2 \theta}}{{2 \times 9.8}}$

Now, solve for $\sin^2 \theta$:

$\sin^2 \theta = \dfrac{{3.3 \times 2 \times 9.8}}{{(11)^2}}$

$\sin^2 \theta = \dfrac{{64.68}}{{121}}$

Now, find $\theta$ by taking the inverse sine:

$\theta = \sin^{-1} \sqrt{\dfrac{{64.68}}{{121}}}$

$\theta = \sin^{-1} (\sqrt{0.5345454545})$

$\theta \approx \sin^{-1} \left(0.731126155\right)$

$\theta \approx 46.98088686^\circ$

$\theta \approx 47^\circ \textsf{( in nearest whole number)}$

Therefore, the water balloons must be launched at an angle of approximately $43.78^\circ$ to just graze the ceiling.

Answer 2

Answer:

θ = 47.0° (3 s.f.)

Explanation:

To find the launch angle of the water balloons, we can model the scenario as projectile motion and use SUVAT equations.

When a body is projected through the air with initial velocity (u), at an angle of θ to the horizontal, it will move along a curved path. Therefore, trigonometry can be used to resolve the body's initial velocity (u) into its vertical and horizontal components:

• Horizontal component = u cos θ
• Vertical component of = u sin θ

In this case, the initial velocity of the water balloons is u = 11 m/s, so the vertical and horizontal components of the initial velocity are:

• Horizontal component = 11 cos θ
• Vertical component of = 11 sin θ

For the water balloons to just graze the ceiling, they must reach a maximum height of 3.3 m. At the maximum height, the vertical component of the velocity becomes zero. Therefore, we can resolve vertically, taking up as positive:

$\textsf{Displacement:}\quad s = 3.3 \;\sf m$

$\textsf{Initial velocity:}\quad u = (11 \sin \theta) \;\sf m/s$

$\textsf{Final velocity:}\quad v = 0 \;\sf m/s$

$\textsf{Acceleration:}\quad a = -9.8 \;\sf m/s^2$

Now, substitute the values into the SUVAT equation v² = u² + 2as, and solve for θ:

$\begin{aligned}v^2&=u^2+2as\\\\\implies (0)^2 &= (11\sin \theta)^2 + 2(-9.8)(3.3)\\\\0 &=121\sin^2 \theta -64.68\\\\121\sin^2 \theta &=64.68\\\\\sin^2 \theta &=\dfrac{64.68}{121}\\\\\sin \theta &=\sqrt{\dfrac{64.68}{121}}\\\\\theta &=\sin^{-1}\left(\sqrt{\dfrac{64.68}{121}}\right)\\\\\theta &=46.98088686189...^{\circ}\\\\\theta &=47.0^{\circ}\; \sf (3\;s.f.)\end{aligned}$

Therefore, the water balloons must be launched at an angle of 47.0° (rounded to three significant figures) to just graze the celling.