1. Luzviminda threw a ball upward with an initial velocity of 2000 cm/s and was
ID: 1950998 • Letter: 1
Question
1. Luzviminda threw a ball upward with an initial velocity of 2000 cm/s and was able to catch it before it reached the ground on its return.a. What was its initial velocity after 1 second?
b. What was its displacement in the first second?
c. How long did it take the ball to reach its maximum height?
d. How far was the maximum from the starting point?
3. How long will it take the ball in problem #2 to reach a point 1000cm above Luzviminda's hands on its way down?
2. An anti-aircraft shell is fired vertically upward with a muzzle velocity of 488 m/s. Determine
a. The maximum height it can reach
b. The time it takes to reach this height;
c. The instantaneous velocities at the end of 40 and 60 seconds.
Explanation / Answer
h=h0+v0t+0.5at2
v=v0+at
here we have h0=0,v0=20[m/sec],a=-9.81[m/sec2]
when t=1 we get:
a. v=20-9.81*1=10.19[m/sec] (please note that the initial velocity is 20[m/sec]. this isn't the initial velocity since the initial velocity doesn't change over time, it's an initial condition!)
b. h=0+20*1-0.5*9.81*12=15.095[m]
c. the question is when v=0. We get: 0=20-9.81t --> t2.04[sec]
d.h=0+20*2.04-0.5*9.81*2.04220.34[m]
3. the question is when h=10[m]. We get: 10=0+20t-0.5*9.81t2
the solution is: t1 0.583[sec] and t2 3.494[sec]. So on the way down, the time is the later one, meaning t3.494[sec]
2. h=h0+v0t+0.5at2
v=v0+at
where h0=0, v0=488[m/sec] and a=-9.81[m/sec2]
a. the time it takes the rocket to be at it's maximum height is when it's velocity iszero, meaning:
0=488-9.81t --> t49.7[sec]
h=0+488*49.7-0.5*9.81*49.72=12137.8[m]
b. we've already calculated it to be t=49.7[sec]
c. t=40, v=488-9.81*40=95.6[m/sec]
t=60, v=488-9.81*60=-100.6[m/sec], meaning a velocity downwards.
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