Two physics problems?
Hello, I've become somewhat stumped on these two physics problems that I was assigned for homework.
The first being: "A baseball leaves a bat at 27.0 m/s at angle of 27 degrees above the horizontal. The ball was 2 meters above the ground when it was hit. Using this information, how far will it travel?"
I understand how to do this problem if if didn't say "it was hit at 2 meters above the ground." If that part of the question wasn't present, you could just use the range equation and plug in the values. However, I have no idea how to solve this given the fact that it was hit 2 meters above the ground.
The next is: The city shoots off fireworks with an initial velocity of 245 m/s at an angle of 44 degrees above the horizontal. If they are meant to explode at 1000 m, what time should the fuse be set for?"
There's another snag here. I could use the formula for finding time for reaching its peak, but that formula requires the object reaches its initial height again, but it won't because it exploded. I got an answer of 16 seconds for this one, but I think it's wrong.
Update:Huh? Are you telling me to ask these problems as two different questions on YA, or did your post get cut off?
Comments
"A baseball leaves a bat at 27.0 m/s at angle of 27 degrees above the horizontal. The ball was 2 meters above the ground when it was hit. Using this information, how far will it travel?"
1. Break the problem into 2 parts: First, solve the problem to find the range, assuming that a fielder catches the ball 2 meters above the ground, using your "range" equation.
2. Now, let's find the ADDITIONAL horizontal distance, if the catcher misses and the ball falls to the ground:
We know what the horizontal component of velocity is, thus:
horizontal distance = vt = (27 m/s)(cos 27 deg)t
horizontal distance = (27)(0.891)t = 24.1t
To find t = time to fall 2 meters to the ground:
s = 2 = (27)(sin 27 deg)t + 0.5(9.81 m/s/s)t^2
We place the quadratic equation in standard form, and solve for t:
4.905t^2 + 12.26t - 2 = 0
t = 0.154 seconds, and plugging this into the equation for horizontal distance:
horizontal distance = 24.1t = (24.1)(0.154) = 3.71 meters.
-------------------
--------------------
.
The city shoots off fireworks with an initial velocity of 245 m/s at an angle of 44 degrees above the horizontal. If they are meant to explode at 1000 m, what time should the fuse be set for?"
There's no "snag". However, we're not trying to find the time to reach the highest point for an object traveling a parabola-arc. We are simply solving for the time in this equation:
s = 1000 m = vt - 0.5at^2
v = 245 m/s(sin 44) = (245)(.695) = 170.3 m/s
1000 = 170.3t - 4.905t^2
4.905t^2 - 170.3t + 1000 = 0
t = 7.486 seconds
Note: the quadratic has 2 solutions = 2 points on the arc; but we only want the quicker time.
-------------
.