Answer: Got It!
<em>Explanation: </em>let s = speed at launch
v = 0 at top = s sin 63 - g t
so at top
t = s sin 63/g = .0909 s
h = 13.6 = s sin 63 t - 4.9 t^2
13.6 = .081s^2 - .0405 s^2
s^2 = 336
s = 18.3 m/s
0 0
On Jupiter, C. your weight would increase by a factor of 2.4 . Weight is a product of mass and gravity. Mass does not change dependent upon location.
It's momentum is twice as much.
Of the forces listed I think the force of him diving and sliding across the infield acted on the player.
I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.
I'll tell you how I look at this, although I may be missing something important.
Position = x(t) = 0.5 sin(pt + p/3)
Speed = position' = x'(t) = 0.5 p cos(pt + p/3)
Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)
At (t = 1.0),
x ' '(t) = -0.5 p² sin( 4/3 p )
In order to evaluate this, don't I still have to know what 'p' is ? ?
I don't think it can be evaluated with the information given in the question.