Revision Projectile motion

Revision for TERM1-
Horizontal projectile
Projectile launched at an angle.

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​Projectile motion

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Multiple Choice

What happens to the horizontal velocity of a projectile as it travels along the curved path?

• increases

• decreases

• stays the same

• none of these

Multiple Choice

What happens to the vertical velocity of a projectile as it goes up?

• increases

• decreases

• stays the same

• none of these

Multiple Choice

What is the vertical acceleration of a projectile due to gravity?

• -9.0 m/s

• -9.8 m/s

• -9.8 m/s2

• -9.08 m/s2

Multiple Choice

What is the horizontal acceleration of a projectile?

• 0 $$\frac{m}{s^2}$$  

• -9.8 $$\frac{m}{s^2}$$  

Multiple Choice

A soccer ball is kicked with an initial velocity of 36 m/s at an angle of 50 degrees. What is the vertical acceleration at point C?

• 50 $$\frac{m}{s^2}$$  

• -9.8 $$\frac{m}{s^2}$$

• 0 $$\frac{m}{s^2}$$

• 36 $$\frac{m}{s^2}$$

Multiple Choice

A soccer ball is kicked with an initial velocity of 36 m/s at an angle of 50 degrees. What is the vertical speed at point C?

• 50 m/s  

• -9.8 m/s

• 0 m/s

• 36 m/s

Multiple Choice

The soccer ball has an initial horizontal velocity of 34 m/s at point A. What is the horizontal velocity at point B?

• 24 m/s  

• -9.8 m/s

• 34 m/s

• 0 m/s

Multiple Choice

A golfer wants to hit his golf ball further. At what angle should he launch the golf ball at for it to reach the maximum range?

• 75 $$\degree$$  

• 30 $$\degree$$  

• 45 $$\degree$$  

• 15 $$\degree$$  

Multiple Choice

Which image below accurately represents the vertical and horizontal velocity vectors on a ball thrown into the air at an angle?

Multiple Choice

A projectile is launched with 45m/s at an angle 30^o above the horizontal. What is the V_y when the projectile hits the ground.

• + 22.5 m/s

• + 39 m/s

• - 22.5 m/s

• - 39.5 m/s

Multiple Choice

A basketball is thrown with an initial velocity of 15 m/s at an angle of 30-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the basketball.

• 2.60 s, 21.65 m, 7.5 m

• 1.5 s, 11.48 m, 2.87 m

• 3.90 s, 28.35 m, 12.5 m

• 4.50 s, 31.70 m, 15.0 m

Multiple Choice

A golf ball is hit horizontally off a 25.0-meter high hill and lands a distance of 40.0 meters from the edge of the hill. Determine the initial horizontal velocity of the golf ball.

• 4.0 m/s

• 12.0 m/s

• 18.0 m/s

• 40.0 m/s

Multiple Choice

A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

• 1.23 s, 10.5 m, 2.5 m

• 2.46 s, 21.0 m, 5.0 m

• 1.15 s, 6.9 m, 1.619 m

• 4.92 s, 42.0 m, 10.0 m

Explanation Slide...

The question asks for the time of flight, horizontal distance, and peak height of a football kicked with an initial velocity of 25 m/s at a 45-degree angle. The correct answer is 3.60 s, 45 m, 16 m. To calculate the time of flight, we can use the formula t = 2 * (v * sin(theta)) / g, where v is the initial velocity, theta is the launch angle, and g is the acceleration due to gravity. Substituting the given values, we get t = 2 * (25 * sin(45)) / 9.8 = 3.60 s. The horizontal distance can be calculated using the formula d = v * cos(theta) * t, which gives d = 25 * cos(45) * 3.60 = 45 m. The peak height can be found using the formula h = (v^2 * sin^2(theta)) / (2 * g), which gives h = (25^2 * sin^2(45)) / (2 * 9.8) = 16 m. Therefore, the time of flight is 3.60 s, the horizontal distance is 45 m, and the peak height is 16 m.

Multiple Choice

A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.

• 4.51 s, 25 m, 0 m

• 3.60 s, 45 m, 16 m

• 1.13 s, 12.5 m, 12.5 m

• 0.56 s, 6.25 m, 6.25 m

Explanation Slide...

To determine the initial horizontal velocity of the soccer ball, we can use the formula for horizontal motion: distance = velocity × time. Since the ball is kicked horizontally, the initial vertical velocity is zero. Using the given values of 22.0 meters for the height and 35.0 meters for the distance, we can rearrange the formula to solve for the velocity. The initial horizontal velocity of the soccer ball is calculated to be 16.5 m/s.

Multiple Choice

A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

• 16.5 m/s

• 9.8 m/s

• 22.0 m/s

• 35.0 m/s

Explanation Slide...

The question asks for the time and distance for a pool ball to fall from a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. The correct answer is 0.350 s, 0.84 m. To calculate the time, we can use the formula t = sqrt(2h/g), where h is the height and g is the acceleration due to gravity. Plugging in the values, we get t = sqrt(2*0.60/9.8) = 0.350 s. The horizontal distance can be calculated using the formula d = v*t, where v is the initial velocity and t is the time. Plugging in the values, we get d = 2.4*0.350 = 0.84 m.

Multiple Choice

A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

• 0.122 s, 0.84 m

• 0.350 s, 0.84 m

• 0.60 s, 0.350 m

• 2.4 s, 0.60 m

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