Please help me to solve this problem.
A body is projected vertically upwards with a velocity 20m/s. find the maximum height reached by the body.(g=10m/sec2)
v = Vi + a t
we want the top where v = 0
Vi = +20
a = -10
so
0 = 20 - 10 t
t = 2 seconds to top
h = Vi t + (1/2) a t^2
h = 20 (2) - (1/2)(10) (2)^2
h = 40 - 20
h = 20
20M
To find the maximum height reached by the body, we need to use the kinematic equation for vertical motion:
h = (v^2 - u^2) / (2g)
Where:
h = maximum height reached by the body
v = final velocity (when the body reaches its highest point)
u = initial velocity (when the body is projected upwards)
g = acceleration due to gravity
Given:
u = 20 m/s
g = 10 m/s^2
To find the final velocity, we can use the fact that when the body reaches its highest point, its final velocity is 0 m/s. So we can set v = 0 in the equation.
0 = (v^2 - u^2) / (2g)
Now, we can rearrange the equation to solve for v:
v^2 - u^2 = 0
v^2 = u^2
v = √(u^2)
v = √(20^2)
v = √400
v = 20 m/s
Now that we have the final velocity, we can plug it back into the original equation to find the maximum height:
h = (v^2 - u^2) / (2g)
h = (20^2 - 20^2) / (2 * 10)
h = (400 - 400) / 20
h = 0 / 20
h = 0
Therefore, the maximum height reached by the body is 0 meters.