Question

If a rock is thrown upward on the planet Mars with a velocity of 17 m/s, its height (in meters) after t seconds is given by H = 17t - 1.86t^2.

(a) Find the velocity of the rock after two seconds.

(b) Find the velocity of the rock when t = a.

(c) When will the rock hit the surface? (Round your answer to one decimal place.)

(d) With what velocity will the rock hit the surface?

Answer 1

Equation of motion: s = ut + .5 a t^2, where s = distance, u = initial velocity and a = acceleration

You are given the equation H = 17t - 1.86t^2

So you see that u = 17 and

.5 a = -1.86 therefore a = -3.72

a) v at t=1 is?

v = u + at <- Equation of motion
= 17 -3.72(1)
= 13.28m/s

b) v at t=a is?

v = 17 - 3.72a

c) t when h = 0 is?

H = 17t - 1.86t2
0 = t (17 - 1.86t)

which is true at:

t = 0

and

0 = 17 - 1.86t -> t = 9.14 secs

d) v at t = 9.14?

v = u + at <- Equation of motion
= 17 - 3.72 (9.14)
= -17 m/s

Which makes sense since we're ignoring resistance.

Answer 2

Equation of motion: s = ut + .5 a t^2, where s = distance, u = initial velocity and a = acceleration

You are given the equation H = 17t - 1.86t^2

So you see that u = 17 and

.5 a = -1.86 therefore a = -3.72

a) v at t=1 is?

v = u + at <- Equation of motion
= 17 -3.72(1)
= 13.28m/s

b) v at t=a is?

v = 17 - 3.72a

c) t when h = 0 is?

H = 17t - 1.86t2
0 = t (17 - 1.86t)

which is true at:

t = 0

and

0 = 17 - 1.86t -> t = 9.14 secs

d) v at t = 9.14?

v = u + at <- Equation of motion
= 17 - 3.72 (9.14)
= -17 m/s