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:
[tex]E= \frac{1}{2}mv^2 [/tex]
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:
[tex]E=mgh[/tex]
for the law of conservation of energy, the initial mechanical energy must be equal to the final mechanical energy, so we have
[tex] \frac{1}{2}mv^2 = mgh [/tex]
From which we find the maximum height of the ball:
[tex]h= \frac{v^2}{2g}= \frac{(20 m/s)^2}{2 \cdot 9.81 m/s^2}=20.4 m [/tex]
Therefore, the answer is yes, the ball will reach the top of the tree.
