Question

When a ball is thrown vertically upwards, its height h metres
after t seconds is modelled
by h= 20t-5t^2
Find the maximum height it reaches​

Answer 1

Step-by-step explanation:

apply second equation of motion.

let initial velocity be u and acceleration due to gravity be g=10m/s^2

therefore when ball is thrown vertically upward we get, the height reached in time t is

h=ut -1/2gt^2

= ut - 1/2 * 10 t^2

= ut - 5t^2

now given that height h can be modelled as h= 20t -5t^2

from the two equations, we can compare the coefficients of t and get that initial velocity (u)=20m/s

Upon reaching the maximum height(H) the final velocity (v) of the ball will be equal to 0.

Applying third equation of motion we get,

v^2= u^2 - 2gH (the direction of velocity of the ball and acceleration due to gravity are oppositte each other)

or, 0 = 20^2 -2*10*H

or, H= 400/20 = 20m

ANSWER: Max height is 20 m