Posted by Kayla on Wednesday, September 28, 2011 at 12:56pm.
A ball launched from ground level lands 2.7 s later on a level field 50 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)
A. Find the magnitude m/s
B. Find the direction ° (above the horizontal)

Physics  tchrwill, Wednesday, September 28, 2011 at 1:57pm
For a projectile fired at an angle of µº to the horizontal with a velocity of Vm/s, the maximum height reached derives from h = V^2(sin^2(µ))/29 and the distance traveled parallel to the ground derives from d = V^2(sin(2µ))/g
1Vcos(µ) = 50/2.75 = 18.181818
2Vsin(µ) = 9.8(2.75) = 26.95
3Vsin(µ)/Vcos(µ) = 26.95/18.18 = 1.48239
4(µ) = arctan(1.48239) = 56º
5Therefore, the initial launch velocity is V = 18.1818/cos(56) = 32.51m/s in the direction of 56º above the horizontal.
I hope this has been of some help to you.
Answer This Question
Related Questions
 physics  A ball launched from ground level lands 3 s later on a level field 48 ...
 physics  A ball launched from ground level lands 2.15 s later on a level field ...
 physics  A projectile is launched over level ground at a launch angle of 70o ...
 johnny  A projectile is launched over level ground at a launch angle of 70o ...
 physics  A projectile is launched at an angle of 58.5 degrees above the ...
 Physics  At time t = 0, a projectile is launched from ground level. At t = 2.00...
 Physics  A soccer player kicks a ball from the ground into the goal, which is ...
 Physics  A ball thrown into the air lands on the same horizontal level, 31 m ...
 physics  A projectile is launched at ground level with an initial speed of 56.0...
 Physics  A projectile is launched at ground level with an initial speed of 51.5...
More Related Questions