Ball moving down roof - kinematics

[bluevirgo80]

A ball rolls down a roof that makes an angle of 30 degrees to the horizontal. It rolls off the edge with a speed of 5m/s. The distance to the ground from that point is two stories or 7m. (a.) How long is the ball in the air? (b) How far from the base of the house does it land? (c) What is its speed just before landing?

I figured (a) is 1.2s and (b) is 6 meters (hopefully I'm correct) how do I figure out (c)?
Thanks!

 

[The Bob]

A ball rolls down a roof that makes an angle of 30 degrees to the horizontal. It rolls off the edge with a speed of 5m/s. The distance to the ground from that point is two stories or 7m. (a.) How long is the ball in the air? (b) How far from the base of the house does it land? (c) What is its speed just before landing?

I figured (a) is 1.2s and (b) is 6 meters (hopefully I'm correct) how do I figure out (c)?
Thanks!

Does the roof have a smooth, horizontal surface before it 'rolls off'? I ask because your calculation is not considering the downwards component of the velocity after it has left the roof. If the ball does not roll off horizontally, then I believe the answer is 0.9s. However, if it does roll of horizontally, then you are correct.

If I am correct in assuming you have to consider the angle of the roof then I also think part (b) is incorrect and should be about 4.00 metres from the base fo the house. Again, if I am not correct in assuming the slope matters and that the ball rolls off horizontally, then you are correct again.

The last one, again, is due to components. You know that v = u + at and by setting a = g (or -g depending on your direction) then you just plug in your value for the time and the initial velocity (u) and you have v just before it lands. Again, I reckon it will be the vector due to two components.

I hope that helps a little.

The Bob (2004 ©)

 

[kyle8921]

(a) To calculate the time the ball is in the air, we can use the equation t = √(2h/g), where t is the time, h is the height, and g is the acceleration due to gravity (9.8 m/s²). Plugging in the values, we get t = √(2*7/9.8) = 1.2 seconds.

(b) To calculate the distance from the base of the house, we can use the equation x = v₀t + ½at², where x is the distance, v₀ is the initial velocity, t is the time, and a is the acceleration. Since the ball is rolling off the edge with a speed of 5 m/s, v₀ = 5 m/s. Plugging in the values, we get x = 5*1.2 + ½*-9.8*(1.2)² = 6 meters.

(c) To calculate the speed just before landing, we can use the equation v = v₀ + at, where v is the final velocity, v₀ is the initial velocity, a is the acceleration, and t is the time. Since the ball is accelerating due to gravity, a = -9.8 m/s². Plugging in the values, we get v = 5 + (-9.8*1.2) = -2.6 m/s. However, since the ball is traveling in a downward direction, the final speed will be positive, so we take the absolute value to get the speed just before landing as 2.6 m/s.