Answer: a) 7.56m/s b) 6.245m/s on its way up and -6.245m/s on its way down.
Explanation:
a) Velocity after 2 seconds
H = 15t - 1.86t^2
V(0) = 15 m/s
Now let’s first differentiate h with respect to t , we have
d/dt x h = d/dt (15t - 1.86t^2)
= 15 d/dt x t - 1.86 d/dt x t^2
= 15 x 1 - 1.86(2t) = 15 - 3.72t
t = 2 sec
Therefore
V(t) = 15 - 3.72(2) = 7.56 m/s
b) if height is 25m
H = 15t - 1.86t^2
Therefore h = 25, we have
25 = 15t - 1.86t^2
15t - 1.86t^2 - 25 = 0
1500t - 186t^2 - 2500 = 0
-186t^2 + 1500t - 2500 = 0
Using quadratic equation to find the roots of equation
First root t1 = 2.3535
Second root T2 = 5.71102
Now, at t1,
V(t1) = 15 - 3.72 (2.3535) = 6.245 m/s
The velocity of the rock on its way up is 6.245 m/s
V(T2) = 15 - 3.72 (5.71102) = -6.245 m/s
The velocity of the rock on its way down is -6.245 m/s
