Question

(b) A stone is thrown straight up with velocity 14 m s-1 Calculate
(i) The time taken for the stone to reach the top of its motion
(ii) The time taken for it to reach the ground again
(iii) The maximum height it reaches
(iv) What is the velocity of the stone at height 8.0 m?

Answer 1

your equation is
height = -4.9t^2 + 14t , where t is in seconds and height is in metres.

velocity = -9.8t + 14

i) at the top of its path, velocity = 0
98t = 14
t = 14/9.8 = 10/7

ii) to reach the ground, height = 0
-4.9t^2 + 14t=0
-t(4.9t - 14) = 0
t=0 , which would be the original position
or
t = 14/4.9 or appr 2.86 seconds

iii) at 10/7 seconds
height = -4.9(10/7)^2 + 14(10/7) = 10 m

iv)
at a height of 8 m
-4.9t^2 + 14t = 8
4.9t^2 - 14t + 8 = 0
by the formula
t = 2.067 or .7897a negative
at t = 2.067
velocity = -9.8(2.067) + 14 = -6.26 m/sec (on its way down)
at t = .7897 sec
velocity = 9.8(.7897)+14 = +6.26 m/s (on its way up)

notice the same speed ?