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Semenov [28]
3 years ago
9

A wheelchair moves upward on a 7.1 degree ramp at a speed of 20km/h what is the horizontal velocity

Physics
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

If the wheelchair is up 7.1 ft. In hight the time of flight should be 0.664 seconds and the distance should be 12.108 ft.

Explanation: I divided the displacement by the time and I used the equation Vx = 20 km/m

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V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )  
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At  α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t² 
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m

GOOD LUCK AND HOPE IT HELPS U
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