Question

Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) A baseball is thrown upward from a height of 3 meters with an initial velocity of 10 meters per second. Determine its maximum height. (Round your answer to two decimal places.)

Answer 1

Answer:8.1 m

Explanation:

Given

ball is launched from height of 3 m

initial velocity $u=10 m/s$

considering the ball is thrown vertically upward

Using $v^2-u^2=2 as$

where,

u=initial velocity

v=final Velocity

a=acceleration

s=distance

At maximum height final velocity will be zero

$v^2-u^2=2 gs$

$0-10^2=2(-9.8)\cdot h$

$h=5.10 m$

Therefore maximum height w.r.t ground is $3+5.10=8.10 m$