Matt shoots the ball with an initial velocity of 20 m/s at an angle of 53 degree
ID: 1656295 • Letter: M
Question
Matt shoots the ball with an initial velocity of 20 m/s at an angle of 53 degrees above the horizontal. Three seconds later the ball swooshes through the hoop. His friend Mo, is waiting directly under the basket. You may neglect air resistance and assume that g= 10 m/s^2.
a) find the components of the initial velocity vector
b) how far is Matt ffrom Mo when he shoots the ball
c) what is the y-component of the ball's velocity when it goes through the hoop.
d) what is the speed of the ball when it goes through the hoop
e) how much higher is the hoop than the release point of the ball
Explanation / Answer
Given,
Initial velocity of ball, vo = 20 m/s
Angle, theta = 53°
A) x-component, vx = 20 x cos53° = 12.03 m/s
y-component, vy = 20 x sin53° = 15.97 m/s
B) Range, R = u2 sin(2 x theta)/g = 202 x sin106°/10= 38.45 m
C) As time, t = 3 sec
vy = voy - (g x t)= 15.97 - (10 x 3) = 14.03 m/s (downward direction)
D) V = (14.032 +12.032) = 18.48m/s
E) y = (voyt)-(0.5 x g x t2) = (15.97 x 3)-(0.5 x 10 x 9) = 2.91 m