A soccer ball was hit and moved upward at a 50.0° angle with an initial velocity of 40.0 m/s.
a. Find the length of time of flight.
b. What is the horizontal distance reached by the soccer ball?
c. What is the maximum vertical distance reached by the soccer ball?
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I will solve the problem as physicists and the Physics students do it . . .
Having given the vector v as v = (40 m/s, 50°), calculate first its horizontal and verical components
= 40*cos(50°) = 40*0.643 = 25.725 m/s;
= 40*sin(50°) = 40*0.765 = 30.630 m/s.
(a) the length of time flight = = = 6.245 seconds.
Here g = 9.81 m/s^2 is the gravity acceleration.
(b) the horizontal distance = = 25.725*6.245 = 160.65 meters.
(c) the maximum vertical distance height = = = 47.82 meters.
Solved.
All questions are answered.