The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s
Distance = (3*5) + (1/2*1*5^2)= 15+12.5= 27.5m
So E = 2x10^-3W/m^2*(π*(3.0x10^-3m)^2)*1min*60s... = 3.4x10^-6J
Answer:
d = 19.796m
Explanation:
Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.
So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.
Next, you look for a kinematic equation that has these parameters and the one you should choose is:
Now by substituting values in, we get
d = 19.796m
A) The acceleration is due to gravity at any given point if you look at it vertically, so
.
b)
, so
. We use
and then the final speed must be 0 because it stops at the highest point. So
. Solve for
and you get
c)
, and then we plug the values:
and we already have the time from "b)", so
; then we just rearrange it
and finally