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A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landing is angled at 60 degree with the roof.

a) find the horizontal distance d it travels. b) what is the magnitude of the balls initial velocity?
c) what is the angle (relative to the horizontal) of the balls initial velocity?

User Mrfour
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2 Answers

1 vote

Final answer:

The horizontal distance can be calculated using the time of flight and horizontal velocity, the magnitude of the ball's initial velocity is obtained via projectile motion equations, and the angle of release is found through the ratio of the vertical to horizontal velocity components.

Step-by-step explanation:

To find the horizontal distance (d) the ball travels, we can use the horizontal velocity and the time it is in air. Since the ball lands 4 seconds later and the horizontal velocity (vx) remains constant (as we ignore air resistance), we can find d using d = vx × t.

The magnitude of the ball's initial velocity can be found by breaking it into horizontal (vx) and vertical (vy) components and using the kinematic equations for projectile motion. We would use the given height of 20 meters, angle of the path just before landing (60 degrees with the roof, which may need to be adjusted depending on the roof's angle with the horizontal), and time of 4 seconds to find the initial vertical speed, and then calculate the initial velocity vector.

The angle of the ball's initial velocity relative to the horizontal can also be determined through the components of the initial velocity vector, using trigonometric relations such as tan(θ) = vy/vx, where θ is the angle we are trying to find.

User Magicgregz
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3 votes
c) what is the angle (relative to the horizontal) of the balls initial velocity?
User ChRyNaN
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