Question

A ball rolls down a roof that makes an angle of 30.0 degrees to the horizontal. It rolls off the edge of the roof with a speed of 5.00 meters per second. The distance straight down to the ground from the edge of the roof is 7.00 meters. a) How long is the ball in the air

Answer 1

The ball is in the air for approximately 1.19 seconds when it rolls off a roof making a 30-degree angle to the horizontal and falling 7 meters to the ground.

Step-by-step explanation:

If a ball rolls off a roof that makes an angle of 30.0 degrees to the horizontal with a speed of 5.00 meters per second, and if the distance to the ground from the roof's edge is 7.00 meters, we can find out how long the ball is in the air. To solve this, we need to consider only the vertical motion since the horizontal and vertical motions are independent. The vertical motion can be described by the equation of free fall: y = vyt + (1/2)gt2, where y is the vertical distance, t is the time in the air, vy is the vertical component of the velocity (which is 0 in this case since the ball rolls off horizontally), and g is the acceleration due to gravity (9.8 m/s2)

Using the vertical distance of 7.00 meters and rearranging the equation for t, we get:

t = sqrt((2y)/g)

Plugging in the numbers:

t = sqrt((2 * 7.00 m) / (9.8 m/s2))

t ≈ 1.19 seconds

Answer 2

0.97s

Step-by-step explanation:

First of all solving the second equation of motion in the vertical direction we have

Vy= Vosinစ

= 5m x sin 30= 2.5m/s

So solving for t time in second equation of motion we will have

t= - Vy +or- √(Vy²- 4{g/2)- h)/ (2g/2)

Substituting 7m for h

We have

t = -2.5+or- √ 2.5²-4(4.9)-7/9.3

We have

t= 0.97s as the positive rational answer