This is from a Capa Physics problem and no one in my class can get the right ans

Question

This is from a Capa Physics problem and no one in my class can get the right answer, please help! (there are 4 parts to it)

1. A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 69.7 m across. If he desires a 2.8-second flight time, what is the correct angle for his launch ramp (deg)?

2. What is his correct launch speed?

3. What is the correct angle for his landing ramp (give a positive angle below the horizontal)?

4. What is his predicted landing velocity. (Neglect air resistance.)

This is from a Capa Physics problem and no one in my class can get the right answer, please help! (there are 4 parts to it) 1. A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 69.7 m across. If he desires a 2.8 - second flight time, what is the correct angle for his launch ramp (deg)? 2. What is his correct launch speed? 3. What is the correct angle for his landing ramp (give a positive angle below the horizontal)? 4. What is his predicted landing velocity. (Neglect air resistance.)

Explanation / Answer

known
y0 = 0 m
y = -15m
x = 69.7 m
x0 = 0m
t = 2.8s

1. find

x = x0 +v0xt

69.7 m = 0 m + v0x(2.8 s)

v0x = 24.9 m/s

y = y0 + v0y t - 0.5 gt2

-15 m = 0m + v0y (2.8s) - 0.5 (9.8m/s2)(2.8s)2

v0y = 8.36 m/s

so

= tan-1 (v0y/v0x)

= 18.6°

2

v0 = sqrt(v0x2 + v0y2)

v0 = sqrt((24.9m/s)2 + (8.36m/s)2)

v0 = 26.3 m/s

3

vx = v0x

vx = 24.9 m/s

vy = v0y - at

vy = 8.36 m/s - (9.8m/s2) (2.8 s)

vy = -19.0 m/s

= tan-1(vy/vx)

= tan-1(-19.0m/s / 24.9 m/s)

= -37.3°

so = 37.3° below the horizontal

4

v = sqrt(vx2 + vy2)

v = sqrt( (24.9m/s)2 + (-19.0 m/s)2)

v = 31.3 m/s

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