Question 192340: I'm having problems with this type of question
Find the height, of a ball, above ground using the following information.
wait 3 seconds, velocity is 45 ft/sec, initial height above ground is 10 ft
wait 2 seconds, velocity is 95 ft/sec, initial height above ground is 50 ft
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! This is a physics problem under "projectile motion"
Sometimes it is found in algebra under quadratic topics as a actual application example.
For today, we are asked how high the projectile will go, in a given time, with an initial velocity. The velocity provides the upward energy, and gravity provides the downward energy.
Usually, the angle of the initial velocity is important, but for today we will assume straight up.
If not, let me know.
The basic eqn is:
delta y = v*t - (1/2) (32.2) t^2
where :
delta y is change in height. Note as time progresses, the projectile goes up, reaches a peak, and then comes down.
v is initial velocity. units in this case are ft per sec or ft/sec
t is elapsed time. units are sec
32.2 is gravity. we are using English today. be careful of metric. english units are 32.2 ft per sec squared, or 32.2 ft/sec^2 (metric is 9.8 m/sec^2)
1/2 is a constant to get average acceleration
1st problem
y= vt-1/2 (32.2) t^2
= (45) (3) - 1/2 (32.2) (3^2)
= 135 - 144.9
= (-9.9) ft
Note we started at 10 ft above ground, therefore 10-9.9 = (.1) ft above ground
This is not so nice a first problem. The projectile has been in the air long enough to go fully up, stop, and now decend to just above ground height.
Additional calc shows that the projectile goes up to a peak of about 31.4 ft in 1.4 sec before starting down.
2nd problem
y=vt-16.1t^2 = 95*2 -16.1 * 2^2= 190-64.4=125.6 ft
Again, initial height was 50 ft, therefore final position is 50+125.6 =175.6 ft
additional calc shows peak at about 3 sec of about 140 ft
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