Question

A cannon fires a cannonball with a muzzle velocity of 50 m/sec at an angle of 40°.
What Is the fligjt Time?
What Is the cannonball maximum altitude?
What Is the cannon's range at tjis angle?

Answer 1

Answer on Question 71526, Physics, Mechanics, Relativity

Question:

A cannon fires a cannonball with a muzzle velocity of $50\ m/s$ at an angle of $40{}^{\circ}$.

a) What is the flight time?

b) What is the cannonball maximum altitude?

c) What is the cannon's range at this angle?

Solution:

a) Let's first find the projections of the initial velocity of the cannonball on axis $x$ and $y$:

$$v_{0x} = v_{0}\cos\alpha = 50\ \frac{m}{s} \cdot \cos 40{}^{\circ} = 38.3\ \frac{m}{s},$$ $$v_{0y} = v_{0}\sin\alpha = 50\ \frac{m}{s} \cdot \sin 40{}^{\circ} = 32.14\ \frac{m}{s}.$$

Let's consider the motion of the cannonball in the vertical direction. We can find the time $t_{rise}$ that the cannonball need to reach the maximum altitude from the kinematic equation:

$$v = v_{0y} - g t_{rise},$$

here, $v_{0y}$ is the projection of the initial velocity of the cannonball on axis $y$, $v = 0$ is the velocity of the cannonball at maximum altitude, $g = -9.8\ \frac{m}{s^2}$ is the acceleration due to gravity.

Then, we get:

$$t_{rise} = \frac{v_{0y}}{g} = \frac{32.14\ \frac{m}{s}}{9.8\ \frac{m}{s^2}} = 3.28\ s.$$

Finally, we can find the flight time multiply $t_{rise}$ by 2:

$$t_{flight} = 2 \cdot t_{rise} = 2 \cdot 3.28\ s = 6.56\ s.$$

b) We can find cannonball maximum altitude from the equation:

$$h = \frac{1}{2} g t_{rise}^{2} = \frac{1}{2} \cdot 9.8 \frac{m}{s^{2}} \cdot (3.28 \, s)^{2} = 52.7 \, m.$$

c) Let's consider the motion of the cannonball in the horizontal direction. We can find the cannon's range at this angle from the formula:

$$x = v_{0x} \cdot t_{flight} = 38.3 \frac{m}{s} \cdot 6.56 \, s = 251.25 \, m.$$

Answer:

a) $t_{flight} = 6.56 \, s$.

b) $h = 52.7 \, m$.

c) $x = 251.25 \, m$.

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