Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
the curve inclination is increased so that a weight component helps keep the car on track
Explanation:
In the sledging competition these devices go at quite high speeds over 100 km/h, so when reaching the curves the friction force is not enough to keep the car on the track. For this reason, the curve inclination is increased so that a weight component helps keep the car on track.
In general we can solve Newton's second law for this case, with the condition of no friction, it is found that
V² = r g tan θ
Where V is the maximum velocity, r is the radius of the curve a, θ is the angle of the inclination
It's attached by earths gravity that keeps it the same force to stay anchored
