If a rock is thrown vertically upward from the surface of mars with velocity 15 mys, its height after t seconds is h − 15t 2 1.86t 2. (a) what is the velocity of the rock after 2 s? (b) what is the velocity of the rock when its height is 25 m on its way up?

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

the answer is to check all of them! they all apply well at least i think so!

the bal will not change its velocity.