Question

4.3 ground launched projectile
morgan mcmiflen
example
shall is thrown at an angle of 30° above the horizontal. the magnitude of the horizontal component of
t velocity is 13.0 metres per second. the magnitude of the vertical component of the balls initial velocit
n per second. (neglect friction.)
hat is the projectiles total time in air?
hat is the projectiles maximum height?
hat is the projectiles range?
multiple-choice question
what is t_total?
0.76 s
1.3 s
1.53 s
rewatch skip

Explanation:

To solve for the total time \( t_{\text{total}} \) of the projectile's motion, we assume the projectile is launched at an angle of \( 30^\circ \) with a horizontal component of velocity \( v_{x} = 13.0 \, \text{m/s} \). First, we find the initial vertical component of velocity \( v_{y} \) using the tangent of the launch angle:

Step 1: Find \( v_{y} \)

The horizontal component \( v_{x} = v_{0} \cos\theta \) and vertical component \( v_{y} = v_{0} \sin\theta \). The ratio \( \frac{v_{y}}{v_{x}} = \tan\theta \), so:
\( v_{y} = v_{x} \tan\theta \)
For \( \theta = 30^\circ \), \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \):
\( v_{y} = 13.0 \, \text{m/s} \times 0.577 \approx 7.50 \, \text{m/s} \)

Step 2: Time to reach maximum height (\( t_{\text{up}} \))

At maximum height, the vertical velocity \( v_{yf} = 0 \). Using the kinematic equation \( v_{yf} = v_{y} - gt_{\text{up}} \) (where \( g = 9.8 \, \text{m/s}^2 \)):
\( 0 = 7.50 - 9.8 t_{\text{up}} \)
Solving for \( t_{\text{up}} \):
\( t_{\text{up}} = \frac{7.50}{9.8} \approx 0.765 \, \text{s} \)

Step 3: Total time of flight (\( t_{\text{total}} \))

The total time is twice the time to reach maximum height (symmetry of projectile motion):
\( t_{\text{total}} = 2 \times t_{\text{up}} \approx 2 \times 0.765 \approx 1.53 \, \text{s} \)

Answer:

1.53 s (corresponding to the option "1.53 s")