Question

A child throws a baseball upward with an initial velocity of 20 m/s. The child wants to throw the baseball at least as high as the very top of the 20 m tall tree in his backyard. (Assume g = 9.80 m/s2 and no air resistance)
Answers :
we need more information
no
yes

Answer 1

When the ball starts its motion from the ground, its potential energy is zero, so all its mechanical energy is kinetic energy of the motion:
$E= \frac{1}{2}mv^2$
where m is the ball's mass and v its initial velocity, 20 m/s.

When the ball reaches its maximum height, h, its velocity is zero, so its mechanical energy is just gravitational potential energy:
$E=mgh$

for the law of conservation of energy, the initial mechanical energy must be equal to the final mechanical energy, so we have
$\frac{1}{2}mv^2 = mgh$
From which we find the maximum height of the ball:
$h= \frac{v^2}{2g}= \frac{(20 m/s)^2}{2 \cdot 9.81 m/s^2}=20.4 m$

Therefore, the answer is yes, the ball will reach the top of the tree.