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

Question

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

 

(a) find the velocity of the rock after one second

(b) find the velocity of the rock when t=a

(c) when will the rock hit the syrface

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

Answer 1

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]

Answer 2

Hi Julia 

Here is the solution

From the above equation acceleration due to gravity at Mars is 0.98 m/s

(a) Using V = u-gt =10 - 0.98*1

V =9.02 m/s

(b) At  t= a,  V = 10- 1.86 *a/2  m/s

(c) Time to reach the highest point

t= 10*2/1.86 sec

So total time  T =2t = 40/1.86 sec

( d) Velocity at the time of hitting the ground will be same as velocity at t=0 when thrown up 

      which is 10 m/s