Question

A tennis ball on Mars, where the acceleration due to gravity is 0.379g and air resistance is negligible, is hit directly upward and returns to the same level 8.5 s later. How fast was it moving just after being hit?

Answer 1

I expect the speed to be 0, s = d/t. The distance moved is virtually nothing and the time is virtually nothing. About 0.

Answer 2

the displacement is 0, t = 8.5 secs and a = 0.379 g

are we talking g = 32 ft/s or 9.8 m/s ?

Answer 3

9.8 or 9.81.

Answer 4

Shouldn't the time be 0? The time it is hit is when the motion begins.

Answer 5

AFAICT.

Answer 6

thats not important anyway let speed = v

so
s = ut + 0.5at^2

0 = 8.5u - 0.5 *0.379* 9.8 * 8.5^2

where u = initial speed

solve for u

Answer 7

* speed is u not v

Answer 8

15.7853

Answer 9

Why did my way not work? We're not hitting the ball at 8.5 seconds.

Answer 10

you cant make an equation when t = 0 and after 8.5 seconds the ball will be travelling at same speed as the intial speed ( though in opposite direction)

Answer 11

15.7853 is correct

Answer 12

So the 8.5 seconds is a way to circumvent the problems arising with plugging in 0 for t?

Answer 13

yea - you could say that

Answer 14

Thank you.

Answer 15

yw

Answer 16

v at any time is not d/t, but it is the derivative of d with respect to t, or in other words, you have to take infinitesimal distance travelled and divide it by infinitesimal time taken.