
b) Finding total distance :
Distance travelled from 0 s to 5 s :
Distance travelled from 5 s to 10 s :
Distance travelled from 10 s to 15 s :
- 150 + 1/2 x 5 x 5
- 150 + 12.5 = 162.5 m
Distance travelled from 15 s to 20 s :
- 36 x 5 + 1/2 x 5 x 4
- 180 + 10 = 190 m
Distance travelled from 20 s to 25 s :
- 45 x 5 + 1/2 x 5 x 5
- 225 + 12.5 = 237.5 m
Distance travelled from 25 s to 30 s :
- 40 x 5 + 1/2 x 10 x 5
- 200 + 25 = 225 m
Distance travelled from 30 s to 35 s :
- 20 x 5 + 1/2 x 20 x 5
- 100 + 50 = 150 m
Distance travelled from 35 s to 40 s :
Total = 75 + 150 + 162.5 + 190 + 237.5 + 225 + 150 + 50
Total = 1240 m
c) velocity at t = 15 s
d) average velocity
- 0 m/s (as displacement is equal to 0)
e) average speed
f) Part d uses displacement whereas part e uses distance
Using the given formula with v0=56 ft/s and h=40 ft
h = -16t2 + v0t
40 = -16t2 + 56t
16t2 - 56t + 40 = 0
Solving the quadratic equation:
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32
We have two possible solutions
t1 = (56+24)/32 = 2.5
t2 = (56-24)/32 = 1
So initially the ball reach a height of 40 ft in 1 second.
So you work hard and you do well i’m getting good grades or doing something better.