Free fall word problem

Question

A ball is thrown upward with an initial velocity of 19.6m/s.what is the max height the ball will reach before falling back down?

 

Just want to check if my answer is correct, I ended up with 19.6m

Answer 1

h(t) = h₀ + v₀t - (1/2)gt²

 

h₀ = 0

 

v₀ = 19.6 m/s

 

g = 9.8 m/s²

 

h(t) = 19.6t - 4.9t²

 

v(t) = v₀ - gt = 19.6 - 9.8t

 

Maximum height is reached when v = 0 and the ball begins to fall

 

0 = 19.6 - 9.8t ⇒ t = 19.6/9.8 s = 2s

 

Plug t = 2s into h(t) and get

 

h_max = h(2) = 19.6(2) - 4.9(2)² = 19.6 m

 

Your result is correct.

Answer 2

You could also work the problem in one step by determining which among the kinematic quantities (displacement, initial velocity, final velocity, acceleration, and time) you know in this situation, and which you want to find.  Typically, if you want to calculate one, you need to know at least three others.  In this case:

 

to find:  displacement (and then height from that)

know:  acceleration (always g, downward, for vertical projectile motion), initial velocity (= +19.6 m/s), and final velocity ( = 0 at maximum height)

 

Then, use the kinematic equation that involves these four quantities:

 

v² = v_o²+2a(x-x_o)

 

where

 

v = final velocity

v_o = initial velocity

a = acceleration

(x-x_o) = displacement (with x = final position (height( and x_o = initial position (height))

 

If we define up as positive and down as negative, we then have:

 

0² = (19.6 m/s)²+2(-9.8 m/s²)(x-x_o)

 

Solve for the quantity (x-x_o) and then solve for x, taking x_o (initial position at the ground) = 0.

 

This gives:  (x-x_o) = 19.6 m, meaning x (final height) = 19.6 m if x_o = 0.  

 

Just another way to approach the same problem, as there often is in physics.