Question

$y=-5t^{3}+15t$

a) find  $y_{max}$

b) time at  $y=y_{max}$

c) acceleration at  $y=y_{max}$

d) velocity just before it reaches ground level

e) distance traveled when  $y=0$

f) time at  $y=0$ (not  $t=0$ solution)

Answer 1

Given

a) To find maximum value of y, we need to set :

So at t = 1s, y = y_max = -5(1)³ + 15(1) = 10 m

b) Time at y = y_max = 1s (as found from above)

c) To find acceleration we need:

So acceleration at y = y_{max =}

d) It takes 1 s to reach y_max, so it takes another 1 s to reach the ground level

so total time = 1 + 1 = 2s

velocity just before it reaches ground level =

e) Total distance travelled = 2 x y_max = 2 x 10 = 20m

f) time at  $y=0$ (not  $t=0$ solution) = 2 s ( as found in (d))