A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m hig
h, forming a 1.1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ
=53.0 degrees above the horizontal at a point d=24.0m from the base of the building wall. The ball takes 2.20s to reach a point vertically above the wall.
a) Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.)
b) Find the vertical distance by which the ball clears the wall.
c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that
t=distance /speed
Hence the speed at which ball was lunched is 18.13m/s
b. from the equation
the vertical distance at which the ball clears the wall is
y=8.14-7.3=0.84m
c. the time it takes the ball to reach the 6.2m vertically
<span>Epsilon
zero is permittivity of free space means how much air or vacuum permits
electric field to travel from one charge to other.It is constant in the coulomb
law. This allow Gauss's a lot easier to solve rather than using K</span>
A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force