A football is kicked at an 50 degree angle to the horizontal , travels a horizontal distance of 20m before hitting the ground. what is the initial speed?
let u be the initial velocity , devide it into two component x and y the angle is 50 degree given so u in x direction = u cos50 in y direction = u sine50 time period = t= (2 u sine50)/g and horizontal distance = u cos50 t ( acceleration in x direction is zero)by putting the value of t in above eqn yuou will get s=ucos50 (2 u sine50)/g 20 g = 2 u^2 cos50 * sine50 from here u can get the value of u which is initial velocity....
Answer:
Initial speed of football=14.11m/s
Explanation:
Range is the horizontal distance travelled. Therefore using formular for Range,R to determine the speed.
R=V^2Sin2theta/g
Given: R=20m
Theta=50°
20= V^2 Sin(2×50)°/ 9.8
Cross multiply
20×9.8 = V^2 Sin100°
196= V^2×0.9848
V=Sqrt(196/0.9848)
V= 14.11m/s