Science: Projectile Motion
Analyzing motion in two dimensions under gravity
Science: Projectile Motion
Analyzing motion in two dimensions under gravity
Science - Grade 9-12
- 1
A ball is rolled horizontally off a table that is 1.2 meters high. It lands 2.4 meters from the base of the table. Find the time the ball is in the air.
For a horizontal launch, use the vertical drop to find the time.
The ball is in the air for about 0.49 seconds. Using vertical motion, t = square root of (2h/g) = square root of (2 x 1.2 / 9.8), which is about 0.49 s. - 2
A ball is rolled horizontally off a table that is 1.2 meters high and lands 2.4 meters away. Find its horizontal speed as it leaves the table.
Its horizontal speed is about 4.9 m/s. First find the time from the vertical motion, about 0.49 s, then use v = d/t = 2.4 / 0.49. - 3
A soccer ball is kicked at 20 m/s at an angle of 30 degrees above the horizontal. Find the horizontal component of its initial velocity.
Cosine gives the horizontal component.
The horizontal component is about 17.3 m/s. Use vx = v cos(theta) = 20 cos(30 degrees). - 4
A soccer ball is kicked at 20 m/s at an angle of 30 degrees above the horizontal. Find the vertical component of its initial velocity.
The vertical component is 10.0 m/s. Use vy = v sin(theta) = 20 sin(30 degrees). - 5
A projectile is launched straight upward with an initial speed of 24.5 m/s. How long does it take to reach its maximum height?
At maximum height, the upward speed has decreased to zero.
It takes 2.5 seconds to reach maximum height. At the top, the vertical velocity is 0, so t = v/g = 24.5 / 9.8 = 2.5 s. - 6
A projectile is launched upward at 24.5 m/s from ground level. What maximum height does it reach?
The projectile reaches a maximum height of about 30.6 meters. Use h = v^2 / (2g) = 24.5^2 / (2 x 9.8). - 7
A toy rocket is launched from ground level at 18 m/s at an angle of 45 degrees. Find its total time of flight.
Find the vertical component first, then double the time to the top.
The total time of flight is about 2.6 seconds. The initial vertical velocity is 18 sin(45 degrees), about 12.7 m/s, and total time is 2vy/g = 2 x 12.7 / 9.8. - 8
A toy rocket is launched from ground level at 18 m/s at an angle of 45 degrees. Find its horizontal range.
The horizontal range is about 33.1 meters. The horizontal speed is 18 cos(45 degrees), about 12.7 m/s, and range is vx times total time, so 12.7 x 2.6 is about 33.1 m. - 9
Two objects are released from the same height at the same time. One is dropped straight down, and the other is launched horizontally. Which one hits the ground first if air resistance is ignored?
Horizontal motion does not change the vertical fall time.
They hit the ground at the same time. Both objects have the same vertical motion because they start with the same vertical velocity and experience the same acceleration due to gravity. - 10
A ball is thrown horizontally at 6.0 m/s from a height of 5.0 meters. How far from the base of the platform does it land?
It lands about 6.1 meters from the base. First find the time using t = square root of (2h/g) = square root of (10/9.8), about 1.01 s. Then use horizontal distance = vt = 6.0 x 1.01. - 11
A projectile is launched at 25 m/s at an angle of 60 degrees. What is its initial vertical velocity?
Use sine for the vertical component.
Its initial vertical velocity is about 21.7 m/s. Use vy = v sin(theta) = 25 sin(60 degrees). - 12
A projectile is launched at 25 m/s at an angle of 60 degrees. What is the maximum height it reaches?
The maximum height is about 24.0 meters. Use the vertical component 21.7 m/s and apply h = vy^2 / (2g) = 21.7^2 / (2 x 9.8). - 13
A student says that a heavier projectile falls faster than a lighter projectile when both are launched the same way and air resistance is ignored. Is the student correct? Explain.
Think about the role of gravity in ideal projectile motion.
The student is not correct. If air resistance is ignored, all projectiles accelerate downward at the same rate, so mass does not change the falling motion. - 14
A projectile is launched from ground level at 30 m/s at an angle of 40 degrees. Find its approximate total time in the air.
The total time in the air is about 3.9 seconds. First find the vertical component, vy = 30 sin(40 degrees), about 19.3 m/s. Then use total time = 2vy/g = 2 x 19.3 / 9.8. - 15
A projectile is launched from ground level at 30 m/s at an angle of 40 degrees. Find its approximate horizontal range.
Range equals horizontal speed times total time in the air.
The horizontal range is about 90.5 meters. The horizontal component is vx = 30 cos(40 degrees), about 23.0 m/s. Multiply by the total time, about 3.9 s, to get a range near 90.5 m.