All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity.
So we solve for v:
m*g*h=(1/2)*m*v², masses cancel out,
g*h=(1/2)*v², we multiply by 2,
2*g*h=v² and take the square root to get v
√(2*g*h)=v, we plug in the numbers and get:
v=9.9 m/s.
So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.
Constant acceleration of plane = 3m/s²
a) Speed of the plane after 4s
Acceleration = speed/time
3m/s² = speed/4s
S = 12m/s
The speed of the plane after 4s is 12m/s.
b) Flight point will be termed as the point the plane got initial speed, u, 20m/s
Find speed after 8s, v
a = 3m/s²
from,
a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>
t
3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>
8
24 = v - 20
v = 44m/s
After 8s the plane would've 44m/s speed.
I think it’s D) all of the above