1. A ball rolls over the edge of a platform with only a horizontal velocity. The
ID: 1315578 • Letter: 1
Question
1. A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground? Neglect air resistance.
2. An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object? Neglect air resistance.
3. A toy rocket is launched vertically from ground level at time t = 0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 61.0 m and acquired an upward velocity of The rocket continues to risewith insignificant air resistance in unpowered flight, reaches maximum height, and falls back to the ground. The maximum height reached by the rocket is closest to
Explanation / Answer
1)
A ball rolls over the edge of a platform with only a horizontal velocity
x=vt
The height of the platform is 1.6 m
s=ut+0.5at2
intial velocity of ball is 0
which is releasing from table
1.6=0*t+0.5at2
1.6=0.5*a*t2
a=g=9.8m/s2
substituting in above equation we will get
t=0.57 seconds
the horizontal range of the ball from the base of the platform is 20 m.
20=0.57*v
v=35.08m/s
2)
An object is dropped from a bridge
both reach the water 20 m below at the same instant
s=ut+0.5at2
intilal velocity=0
20=0.5*9.8*t2
t=2.02 seconds
A second object is thrown downwards 1.0 s later.
it means that A second object meets first object at 1.02 seconds
distance travelled by second object
s=0.5*g*t2
=0.5*9.8*1.022
=5.1m
velocity required by second object to cover remaining distance
1.02v=20-5.1
v=14.6m/s
3)
A toy rocket is launched vertically from ground level at time t = 0 s.
At the instant of engine burnout, the rocket has risen to 61.0 m
v2-u2=2as
here intial velocity=0
a=g=9.8m/s2
v2=2*9.8*61
v=34.57m/s
time for upward lift
34.57/9.8
t=3.52seconds
maximum height
aproximately 60.7m
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