Planet Mars Rock Application

Calculus
Section 2.7







What is the set up for (d)?

 

you made mistake in a)

should be
=

not - 3.72 a - 1.86

 

if rock is thrown upward on the planet Mars with velocity of 10m/s, its height (in meters) after t seconds is given by H=10t-1.86t^2


a. velocity after one second

v(1)=lim(h->0 (H(1+h)-H(1))/h)

v(1)=lim(h->0 (10(1+h)-1.86(1+h)^2-(10(1)-1.86(1^2))/h).........simplify

v(1)=lim(h->0 (6.28-1.86h))

=6.28
so, the velocity is 6.28m/s

b. velocity at t=a

doing same way as fot t=1

you get lim(h->0 (10-3.72a-1.86h)
=10-3.72a

so, at t=a, the velocity is 10-3.72a m/s

c
rock will hit surface when H=0


0=10t-1.86t^2

0=t(10-1.86t)
=> t=0 or t=5.38

d.

using result from part (b)

v(a)=10-3.72a if a=5.38
v(5.38)=10-3.72*5.38
v(5.38)=10-20.0136
v(5.38)=-10.0136 ≈ -10.01

the rock hits the surface with a velocity of ≈ -10.01 m/s

 

I totally messed this up.