10 Questions
A baseball is thrown at an angle of 30° to the horizontal with an initial velocity of 20 m/s. What is the horizontal component of the velocity of the ball when it reaches its maximum height?
10 m/s
A projectile is launched at an angle of 45° to the horizontal with an initial velocity of 30 m/s. What is the range of the projectile?
R = (30 m/s)² sin(2 × 45°) / g = 90 m
A 2 kg block moving at 4 m/s to the right collides with a 3 kg block moving at 2 m/s to the left. If the two blocks stick together after the collision, what is the velocity of the combined blocks?
0.4 m/s to the right
A 5 kg cart moving at 3 m/s to the right collides with a 2 kg cart moving at 2 m/s to the left. If the collision is perfectly elastic, what is the final velocity of the 5 kg cart?
1.5 m/s to the right
A stone is thrown from the top of a building with an initial velocity of 20 m/s at an angle of 60° to the horizontal. What is the time of flight of the stone?
t = 2v₀ sin(θ) / g = 2(20 m/s) sin(60°) / g = 3.46 s
A golfer hits a ball with an initial velocity of 40 m/s at an angle of 40° to the horizontal. What is the maximum height reached by the ball?
Using $v_y = v_0 \sin \theta$, we find the vertical component of velocity as $v_y = 40 \sin 40° = 25.55$ m/s. The maximum height is reached when $v_y = 0$. Using the equation $v_y^2 = v_{0y}^2 - 2gh$, we find the maximum height as $h = \frac{v_{0y}^2}{2g} = \frac{25.55^2}{2 * 9.8} = 33.2$ m.
A cannonball is launched with an initial velocity of 50 m/s at an angle of 70° to the horizontal. Find the horizontal distance traveled by the cannonball.
The horizontal distance traveled is the range of the projectile, which can be found using the equation $R = \frac{v_0^2 \sin 2 \theta}{g} = \frac{50^2 * \sin 140°}{9.8} = 245.45$ m.
A football player kicks a ball with an initial velocity of 25 m/s at an angle of 55° to the horizontal. Find the time of flight of the ball.
The time of flight can be found using the equation $t = \frac{2v_0 \sin \theta}{g} = \frac{2 * 25 * \sin 55°}{9.8} = 4.55$ s.
A basketball player shoots the ball with an initial velocity of 15 m/s at an angle of 35° to the horizontal. What is the horizontal component of the velocity of the ball when it reaches its maximum height?
The horizontal component of velocity remains constant, since there is no acceleration in the horizontal direction. Therefore, the horizontal component of velocity is simply the horizontal component of the initial velocity, which is $v_x = v_0 \cos \theta = 15 \cos 35° = 12.34$ m/s.
A shot-putter throws a shot with an initial velocity of 12 m/s at an angle of 45° to the horizontal. Find the vertical component of the acceleration of the shot.
The vertical component of acceleration is simply the acceleration due to gravity, which is $a_y = -g = -9.8$ m/s².
Test your understanding of projectile motion and conservation of momentum with this collection of problems and solutions. Improve your skills in physics and prepare for exams or assignments.
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