Question

2. An object is launched straight up. The object attains a height of 107 m after 2.50 s.
a. What was the initial velocity?
b. When (i.e. how long after launch) will the object have a speed of 20 m/s on the
way back down?

Answer 1

s=ut+1/2at^2
107=u*2.5- 1/2 * 10* 2.5^2
u= 55.3m/s
v= u+at
when reaching maximum height v=0
0=55.3-10*t1
t1=5.53s
now on the way back v=20m/s
20=10*t2
t2=2
hence t1+t2=7.53
so after 7.73 seconds from the launch the body has a velocity of 20m/s on the way down