Question

I do not understand the answer that was given to me on a post dated Nov. 23.

Please help explain my questions that appear at the end of this post.
Bart
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Posted by Bart on Wednesday, November 23, 2011 at 6:52am.


A baseball is hit with a speed of 27 meters per second at an angle of 45 degrees. It lands on the flat roof of a 13 meter tall building. If the ball was hit when it was 1 meter above the ground, what horizontal distance does it travel before it lands on the building?


A long jumper leaves the ground at 45 degrees above the horizontal and lands 8 meters away.
What is his "takeoff" speed?


A long jumper is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10 meters away horizontally, and 2.5 meters vertically below. If he jumps from the edge of the left bank at a 45 degree angle with the speed calculated in question.
How long or short from the opposite bank will he land?
• physics - drwls, Wednesday, November 23, 2011 at 7:18am
#2: 8.0 m = (Vo^2/g)sin(2A) = Vo^2/g
Solve for Vo

For #1, Vx = Voy = 27/sqrt2 = 19.1 m/s

Y = 1 + Voy*t -(g/2)t^2 = 13
Solve for t. Then, use
X = Vx*t for the horizontal distance

For #3, use what you have learned and try it yourself. I assume you are supposed to use the takeoff speed calculated in the previous question.




For #2 What formula did you use when writing: (Vo^g)sin (2A)? Where did the 2 in the 2A come from?

For #1 Where did the 27/sqrt2 come from?

For #3 I still need help doing this one. I am not yet at the point where I can do it myself.

Answer 1

1. Vo = 27 m/s @ 45 Deg.
Xo = 27cos45 = 19.09 m/s.
Yo = 27sin45 = 19.09 m/s.

Vf = Vo + gt,
t = (Yf - Yo) / g ,
t = Tr = (0 - 19.09 / -9.8 = 1.95 s. =
Rise time = Time to reach max. ht.

hmax = Yo*t + 0.5g*t^2,
hmax = 19.09*1.95 - 4.9(1.95)^2 = 18.59 m.

d=Vo*t + 0.5t^2 = 18.59 - 13 = 5.59 m.
0 + 4.9t^2 = 5.59,
t^2 = 1.14,
t = Tf = 1.07 s. = Fall time = Time to
fall to roof.

T = Tr + Tf = 1.95 + 1.07 = 3.02 s. =
Time in flight.

Dx = Xo*T = 19.09 * 3.02 = 57.7 m. Hor
dist. traveled.

2. Dx = Vo^2*sin(2A)/g = 8 m.
Vo^2*sin(90) / 9.8 = 8,
Vo^2 / 9.8 = 8,
Vo^2 = 78.4,
Vo - 8.85 m/s. = Inital velocity.

3. Dx = (8.85)^2sin(90)/9.8 = 8 m.
d = 10-8 = 2 m. short.