A motorcycle daredevil jumps off a 33.0 ramp at 20.3 m/s. The landing ramp is at the same height, 28.0 m away. What is the height of the motorcycle when it reaches the landing ramp? (Unit = m)

Since the ramp is a 33.0° ramp and the motorcycle daredevil jumps off with a speed of 20.3 m/s, the motorcycle dare devil has a horizontal component of speed u = 20.3cos33.0° m/s and a vertical component of speed v = 20.3sin33.0° m/s.

Now, since the other ramp is d = 28.0 m away, it takes the time it takes the motorcycle dare devil to reach it is t.

Considering motion in the horizontal direction, d = ut.

Thus, t = d/u

= 28.0 m/20.3cos33.0° m/s

= 28.0 m/(20.3 × 0.8387) m/s

= 28.0 m/17.025 m/s

= 1.645 s

Let h be the height of the motorcycle daredevil when it reaches the landing ramp in time, t.

Considering the vertical motion and using h = vt - 1/2gt² where v = vertical velocity of motorcycle daredevil = 20.3sin33.0°, t = time taken to reach landing ramp = 1.645 s and g = acceleration due to gravity = 9.8 m/s² (Note that there is a negative in front of g since it is directed downwards)

So, substituting the values of the variables into the equation, we have

h = vt - 1/2gt²

h = 20.3sin33.0° m/s × 1.645 s - 1/2 × 9.8 m/s² × (1.645 s)²

h = 20.3 × 0.5446 m/s × 1.645 s - 1/2 × 9.8 m/s² × 2.706025 s²

h = 18.187 m - 1/2 × 26.519 m

h = 18.187 m - 13.26 m

h = 4.927

h ≅ 4.93 m

So, the height of the motorcycle daredevil when it reaches the landing ramp is 4.93 m.

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