In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start

Answer:

Explanation:

Applying trigonometry to calculate the slope's height:

sinθ = opposite / hypotenuse

sin 4° = h / 100

h = 100 × sin 4°

h = 6.98 m

Using conservation of energy principles:

Change in potential energy equals change in kinetic energy,

∆PE = ∆KE

mg(hf - hi) = ½ m (vf² - vi²)

The initial speed is:

vi = 9.2 m/s

Initial height is:

hi = 6.98 m

Final height when the sled reaches the ground:

hf = 0 m

We solve for final velocity vf:

mg(hf - hi) = ½ m (vf² - vi²)

Mass cancels out:

g(hf - hi) = ½ (vf² - vi²)

9.81 × (0 - 6.98) = ½ (vf² - 9.2²)

-68.43 = ½ (vf² - 84.64)

Multiplying both sides by 2:

-136.86 = vf² - 84.64

vf² = -136.86 + 84.64 = -52.22 (This is inconsistent; correcting the sign)

Actually, hf - hi = 0 - 6.98 = -6.98, but since the sled is descending, potential energy loss is positive, so flip the signs:

g(hi - hf) = ½ (vf² - vi²)

9.81 × 6.98 = ½ (vf² - 9.2²)

68.43 = ½ (vf² - 84.64)

136.86 = vf² - 84.64

vf² = 136.86 + 84.64 = 221.5

vf = √221.5 = 14.88 m/s

Answer:

v = 14.87 m/s

Explanation:

Given:

Initial velocity, u = 9.2 m/s

Angle θ = 4°

Distance traveled, d = 100 m

Let v be the speed after traveling 100 m.

From the figure:

Height, h = d × sinθ

h = 100 × sin 4°

h = 6.97 m

Using physics equations,

Therefore, v = 14.87 m/s