A stunt pilot is attempting to drop a water balloon from a moving airplane onto a target on the ground. The plane moves at a speed of 75.4 m/s and a 41° above the horizontal when the balloon is released. At the point of release, the plane is at an altitude of 900 m.
(a) How far horizontally, in meters and measured from a point directly below the plane's initial position, will the balloon travel before striking the ground?
(b) At the point just before balloon strikes the ground, what angle does its velocity make with the horizontal? Give your answer as an angle measured in degrees below the horizontal.
Vo = 75.4m/s[41o].
Xo = 75.4*Cos41 = 56.9 m/s.
Yo = 75.4*sin41 = 49.5 m/s.
a. Y^2 = Yo^2 + 2g*h = 49.5^2 + 19.6*900 = 20,090,
Y = 142 m/s. = Final Y-component.
Y = Yo + g*T = 142.
49.5 + 9.8T = 142,
T = 9.44 s. to reach gnd.
d = Xo * T = 56.9 * 9.44 = __ m.
b. Tan A = Y/Xo = 142/56.9.
A = 68.2 degrees.
8
t
a. Correction: Y = Yo + g*T = 142.
-49.5 + 9.8T = 142,
T = 19.5 s. to reach gnd.
d = Xo * T = 56.9 * 19.5 = 1, 112 m.
To solve this problem, we need to break it down into two parts: finding the horizontal distance the balloon travels before striking the ground, and determining the angle its velocity makes with the horizontal at that point.
(a) To find the horizontal distance traveled by the balloon, we use the equation of motion:
\[d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2\]
where
- \(d\) is the horizontal distance traveled,
- \(v_i\) is the initial horizontal velocity of the balloon (which is equal to the horizontal component of the plane's velocity),
- \(t\) is the time taken,
- \(a\) is the horizontal acceleration of the balloon.
The initial horizontal velocity (\(v_i\)) can be found using the given speed of the plane (\(75.4 \, \text{m/s}\)) and the release angle (\(41°\)):
\[v_i = v_{\text{plane}} \cdot \cos(\text{release angle})\]
Substituting the given values:
\[v_i = 75.4 \, \text{m/s} \cdot \cos(41°)\]
Next, we need to find the time it takes for the balloon to reach the ground. Since the plane is at an altitude of \(900 \, \text{m}\) when the balloon is released, we can use the equation:
\[900 = v_{\text{plane}} \cdot \sin(\text{release angle}) \cdot t + \frac{1}{2} \cdot g \cdot t^2\]
where
- \(g\) is the acceleration due to gravity.
Solving this equation for \(t\), we can use the positive root since the time cannot be negative.
Once we have the time, we can substitute it back into the equation for \(d\) to find the horizontal distance traveled by the balloon.
(b) To determine the angle the velocity makes with the horizontal just before the balloon strikes the ground, we can use trigonometry:
\[\text{angle} = \tan^{-1}\left(\frac{v_f}{v_h}\right)\]
where
- \(\text{angle}\) is the angle with the horizontal,
- \(v_f\) is the final vertical velocity of the balloon (which is negative, as it is directed downward),
- \(v_h\) is the horizontal velocity of the balloon.
By using the previously calculated values of the initial horizontal velocity (\(v_i\)) and the time taken (\(t\)), we can find the final vertical velocity (\(v_f\)). The horizontal velocity (\(v_h\)) remains the same as the initial horizontal velocity (\(v_i\)).
Using the above equations and the given values, we can solve for both parts of the problem.