Answer:
v = 14.92 m/s
Explanation:
First, make a <u>free body diagram</u> and see the <u>forces in the y-direction</u>.
Use Newton's 2nd Law F = ma to replace ∑F_y with m * a_y.
The acceleration in the y-direction is the centripetal acceleration, a_c = v^2/r.
The <u>normal force is 0</u> because this is where the rollercoaster is not falling off the track yet not touching the track.
The masses cancel out.
<u>Solve for v</u> to find the speed of the rollercoaster at the top of the loop.
The minimum speed the coaster must have at the top of the loop to not fall off the track is <u>14.92 m/s</u>.
Speed of the TRAX is given as 16 m/s so let say it is given as
now our speed towards the front end is given by 5 m/s so this is the relative speed of us with respect to TRAX
let say this speed is given as
now we need to find the speed with respect to someone standing outside the TRAX
so here we need to find the net speed in ground frame and hence we can use the formula of relative speed
so someone outside the TRAX will see our speed as 21 m/s
Answer:
276.62 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)
Equation of motion
<u>Neglecting air drag</u> the velocity of the spherical drop would be 276.62 m/s