Question

A ball is hit at an angle of 36 degrees to the horizontal from 14 feet above the ground. If it hits the ground 26 feet away (in horizontal distance), what was the initial velocity? Assume there is no air resistance.

math - Henry, Friday, April 8, 2011 at 9:59pm

X = hor. = 26Ft.

r(hyp.) = X / cosA = 26 / cos36=32.1Ft.

Y = ver. = 32.1sin36 = 18.9Ft.

d(down) = Vo + 0.5gt^2 = 18.9,
0 + 4.9t^2 = 18.9.
t^2 = 3.86,
t = 1.96s.

Vf^2 = Vo^2 + 2gd = 0,
Vo^2 + 2*(-9.8)18.9 = 0,
Vo^2 - 370.44 = 0,
Vo^2 = 370.44,
Vo(ver.) = 19.2m/s.

Vo = 19.2 / sin36 = 32.7m/s.

Mr.Henry,

would you please explain this answer with a diagram and represtions.

Answer 1

hello arun~! i think i also answered this yesterday,, i'll just re-post my solution here. if there's something you didn't understand, feel free to ask. :)

first we observe that the motion of the ball is projectile and its trajectory (path followed by the ball) is shaped like a parabola. the horizontal distance travelled by the ball is thus given by:
R = (vo)^2 * sin (2*theta) / g
where
R = range or the horizontal distance
vo = initial velocity
theta = angle of release
g = acceleration due to gravity (9.8 m/s^2 = 32 ft/s^2)
substituting,
26 = (vo)^2 * sin(2*36) / 32
26 = (vo)^2 * 0.02972
26/0.02972 = (vo)^2
874.82 = (vo)^2
getting the squareroot of both sides,
vo = 29.58 ft/s

hope this helps~ :)