Question

If a ball is thrown upward on Mars with an initial velocity of 10.0 m/s and lands 1.50 m below its launch point 5.54 seconds after it is thrown, what is the local acceleration due to gravity on Mars?

Answer 1

Answer:H(t) = 10t - 1.86t2

(a)

v(t) = dH/dt = 10 - 3.72t

v(1) = 10 - 3.72(1) = 6.28 m/s

(b)

v(a) = 10 - 3.72a [Julia, what is "a" ?]

(c)

The rock hits the surface when

H(t) = 0

10t - 1.86t2 = t(10 - 1.86t) = 0 ⇒ t = 0 or 10 - 1.86t = 0

The first case is launch, the second is return to ground.

10 - 1.86t = 0

10 = 1.86t

t = 10/1.86 s ≅ 5.3763 s

(d)

v at surface = v at time from part (c).

v(5.3763) = 10 - 3.72(5.3763) ≅ -10 m/s [should be same as launch speed, but in opposite direction]