Vo = 18m/s @ 29deg.
Xo = hor. = 18cos29 = 15.74m/s.
Yo = ver. = 18sin29 = 8.73m/s.
Dh = Vo^2*sin(2A) / g,
Dh = (18)^2*sin58 / 9.8 = 28m = Hor. dist. of ball.
d = Vt,
V = d/t = (28-19) / 1.782 = 5.1m/s. =
Velocity of receiver.
A quarterback throws a football toward a receiver with an initial speed of 18 m/s, at an angle of 29 degrees above the horizontal. At that instant, the receiver is 19 m from the quarterback. With what constant speed should the receiver run in order to catch the football at the level at which it was thrown? The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s.
2 answers
The following steps should follow the Dh calculation:
Tr = (Yf - Yo) / g,
Tr = (0 - 8.73) / -9.8 = 0.891s = Rise
time = Time to reach max height.
Tf = Tr = 0.891s. = Fall time.
t = Tr + Tf = 0.891 + 0.891 = 1.782s =
Time in flight.
Tr = (Yf - Yo) / g,
Tr = (0 - 8.73) / -9.8 = 0.891s = Rise
time = Time to reach max height.
Tf = Tr = 0.891s. = Fall time.
t = Tr + Tf = 0.891 + 0.891 = 1.782s =
Time in flight.
0 answers