Asked by Eric
You and your friend are practicing pass plays with a football. You throw the football at a 35.0° angle above the horizontal at 19.0 m/s. Your friend starts right next to you, and he moves down the field directly away from you at 5.50 m/s. How long after he starts running should you throw the football? Assume he catches the ball at the same height at which you throw it.
Answers
Answered by
Henry
Vo = (19m/s,35 deg.).
Xo = hor.= 19cos35 = 15.56m/s.
Yo = ver.= 19sin35 = 10.90m/s.
h = (Vf^2 - Yo^2) / 2g,
h = (0 - (10.9)^2) / -19.6 = 6.06m.
t(up) = (Vf - Yo) / g,
t(up) = (0 - 10.9) / -9.8 = 1.11s.
t(dn) = t(up) = 1.11s.
T = t(up) + t(dn) = 1.1 + 1.11 = 2.22s
= Time in flight.
d = Xo*T = 15.56 * 2.22 = 34.5m = hor
distance traveled.
t = d/V = 34.5 / 5.5 = 6.27s
td + T = 6.27s,
td = 6.27 - T = 6.27 - 2.22 = 4.07s.
after the receiver starts running, the
ball should be thrown.
Xo = hor.= 19cos35 = 15.56m/s.
Yo = ver.= 19sin35 = 10.90m/s.
h = (Vf^2 - Yo^2) / 2g,
h = (0 - (10.9)^2) / -19.6 = 6.06m.
t(up) = (Vf - Yo) / g,
t(up) = (0 - 10.9) / -9.8 = 1.11s.
t(dn) = t(up) = 1.11s.
T = t(up) + t(dn) = 1.1 + 1.11 = 2.22s
= Time in flight.
d = Xo*T = 15.56 * 2.22 = 34.5m = hor
distance traveled.
t = d/V = 34.5 / 5.5 = 6.27s
td + T = 6.27s,
td = 6.27 - T = 6.27 - 2.22 = 4.07s.
after the receiver starts running, the
ball should be thrown.
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