d) What is the horizontal displacement of the rocket 2. A physics professor did

Question

d) What is the horizontal displacement of the rocket 2. A physics professor did daredevil stunts in his spare time. His last on a motorcycle (See 15.0 m stunt was an attempt to jump across a river adjacent Figure). The takeoff ramp was inclined at 53.0, the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of e ramp. The river itself was 100 m below the ramp. You can ignore ar 530 100 resistance a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? b) If his speed was only half the value found in a), where did he land? A dog running in an open field has components of velocity v,-2.6 mis and v,--1.8 mis at t, For the time interval from t 10.0 s to t- 20.0 s, the average acceleration of the dog has magnitude 0.45 m/'s and direction 31.0 measured from the +x-axis toward the +y-axis. At t20.09. 3. 1.8 mis at t, 10.0 s. a) b) c) what are the x- and y-components of the dog's velocity?, and what are the magnitude and direction of the dog's velocity? Sketch the velocity vectors at t, and to. How do these two vectors differ? 4. A snowball rolls of a barn roof that slopes downward at an angle of 40 o 7.00 m/s (See adjacent Figure). The edge of the roof is 14.0 m above the ground and the snowball has a speed of 7.00 m/s as it rolls off the roof. Ignore air resistance. a) How far from the edge of the barn does the snowball strike the 14.0 m ground if it does not strike anything else while falling? b) Draw x-t, y t, vt and vt graphs for the motion in part a). c) A man 1.9 m tall is standing 4.0 m from the edge of the barn. Will he be hit by the snowball? 4.0 m

Explanation / Answer

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2.

a)

Vox=VoCos53

Voy=Vosin53

time taken to travel 40 m is

t=x/Vox =40/VoCos53

From

Y=Yo+Voyt-(1/2)gt2

-15=0+(Vosin53)*(40/VoCos53)-(1/2)*9.81*(40/VoCos53)2

Vo=17.84 m/s

b)

Now Vo=17.84/2=8.92 m/s

=>-100=(8.92sin53)*t-(1/2)*9.81*t2

4.905t2-7.12t-100=0

t=5.3 s

Horizontal distance

X=(8.92cos53)*5.3

X=28.45 m

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