Answer to Question #255748 in Mechanics | Relativity for yass

Question

A person launches a ball up. The ball left his hand from a height of 1m above the ground level at an initial 

velocity of 2m/s upward. 1. Sketch the situation of this problem 2. How long the ball is in the air from the instant it left the man’s hand to the moment it hits the 

ground? 3. How high it goes? 4. What is the ball velocity just before it hits the ground? 

Ignore air resistance.

Answer 1

height:

$h=h_0+v_0t-gt^2/2$

we have:

$h_0=1$ m

$v_0=2$ m/s

1.

graph h(t):

2.

$1+2t-4.9t^2=0$

$t=\frac{2+\sqrt{4+4\cdot4.9}}{9.8}=0.7$ s

3.

$h'(t)=2-9.8t=0$

$t=2/9.8=0.2$ s

$h_{max}=h(0.2)=1+2\cdot0.2-4.9\cdot0.2^2=1.2$ m

4.

$v(0.7)=|h'(0.7)|=|2-9.8\cdot0.7|=4.86$ m/s