Answer:
10.2 metres
Explanation:
Given that a ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum height it will reach?
Solution
To get the maximum height, let us use the formula
V^2 = U^2 sin^2ø - 2gH
At maximum height V = 0
U^2 sin^2ø = 2gH
Substitute all the parameters into the formula
20^2 ( sin 45 )^2 = 2 × 9.8 × H
400 × 0.5 = 19.6 H
Make H the subject of formula
H = 200 / 19.6
H = 10.204 metres.
Therefore, the maximum height reached by the projected ball is 10.2 metres.
When you talk about rate, you will expect that it will be in terms of a time unit. It measures how fast it is going. So, you would expect that the denominator is in time units. For the movement, you can measure this with either distance or velocity.
So, for the first variety, you would need distance and time to measure the rate of how far you go at a certain time. It is also called as velocity. For the second variety, you would need velocity and time to measure the rate of how fast you are going at a certain interval. It is also called as acceleration.
Π = c/ f
π = 3×10^8 / 1.5×10^13
π = 3÷1.5 ×10^(8-13)
π = 2×10^-5 m
