Answer:
38.8 m
Explanation:
h = -(21 + 1.75) = - 22.75 m
g = - 9.8 m /s^2
Ux = 14.004 m/s
Uy = + 5.376 m/s
Let the ball hits the ground in time t and at a distance d from the base of hill.
Use second equation of motion
h = Uyt + 0.5 at^2
- 22.75 = 5.376 t - 0.5 x 9.8 t^2
4.9 t^2 - 5.376 t - 22.75 = 0
By solving
t = 2.77 second
So, horizontal distance
d = Ux t
d = 14.004 x 2.77 = 38.8 m
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
Answer:
Explanation:
As per energy conservation we can say that energy stored in the spring at the position of maximum compression must be equal to the kinetic energy of bullet and block system
so here we have
here we know that
k = 205 N/m
x = 35 cm
now by momentum conservation we know that
now plug in all values in it
now from above equation
by solving above equation we have