posted by Bart on .
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?
#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.