Question

what is the velocity of a golf ball traveling a distance of 21 feet? without wind being a factor and angle at 15 degrees 4ft above ground

Answer 1

let

y = 4 ft

= 4*0.3048

= 1.2192 m

x = 21 ft

= 21*0.3048

= 6.40 m

let v is the initial velocity

vx = v*cos(15)
vy = v*sin(15)

let t is the time taken.

x = vx*t

6.40 = v*cos(15)*t

t = 6.4/(v*cos(15))

now in y-direction

-y = vy*t - (1/2)*g*t^2

-1.2192 = v*sin(15)*6.4/(v*cos(15)) - (1/2)*9.8*(6.4/(v*cos(15)))^2

-1.2192 = 6.4*sin(15) - 4.9*(6.4/(v*cos(15)))^2

on solving the above equation we get,

==> v = 8.65 m/s or 28.4 ft/s <<<<<<<-----Answer