1:18 PM T o 100% ooooo T-Mobile LTE session.masteringphysics.com Problem 3.50 Ma

Question

1:18 PM T o 100% ooooo T-Mobile LTE session.masteringphysics.com Problem 3.50 MasteringPhysics Part A With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is l 5 m above the cars and the horizontal distance he must clear is 22 m. Express your answer using two significant figures. m/s vmin E Give Up Submit Part B If the ramp is now tilted upward, so that "takeoff angle" is 8.0 above the horizontal, what is the new minimum speed? 22 m Must clear this point! 1.5 m Tap image to zoom Express your answer using two significant figures. m/s vmin E

Explanation / Answer

a). horizontal distance travelled = V*t
=> V*t = 22

vertical distance travelled = u*t + 0.5*g*t^2
=> 1.5 = 0 + 0.5*9.8*t^2
=> 1.5 = 4.9*(22/V)^2
=> 0.306 = 484/V^2
=> V = 39.77 m/s


b).

horizontal velocity = v*cos8
vertical velocity = v*sin8

horizontal distance travelled = (v*cos8)*t
=>(v*cos9)*t = 22


also,

vertical distance travelled = u*t + 0.5*g*t^2
1.5 = (-v*sin8)*t + 0.5*9.8*t^2
=> 1.5 = (-v*sin8)*[22/(v*cos8)] + 4.9* [22/(v*cos8 )]^2
=> 1.5 = -3.484 + 2431.09/v^2
=> v = 22.09 m/s

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