2) The track for a racing event was designed so that riders jump off the slope at 37 degrees from a height of 1 m. During a race it was observed that the rider remained in mid air for 1.5 seconds. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground and the maximum height he attains. Neglect the size of the bike and rider.

3.277 m

Explanation:

Given :

Maximum Height (Hmax) = (u²sin²θ) / 2g

Xv = Xh + Uv * t + 0.5gt²

Xv and Xh are vertical and horizontal distances

-1 = 0 + sin37 * 1.5 Uv + 0.5*-9.8*1.5^2

-1 = 0 + 0.903Uv - 11.025

-1 + 11.025 = 0.903Uv

10.025 = 0.903Uv

Uv = 10.025 / 0.903

Uv = 11.10 m/s

Hmax = 1 + (u²sin²θ) / 2g

= (11.10^2 * (sin37)^2) / 2*9.8

= 44.624360 / 19.6

= 2.277

Hmax = 1 + 2.277

Hmax = 3.277 m

correct it is c.) b b