A ball thrown by a boy in the street is caught 5 s later by another boy on the balcony of the house 10 m away and 3 m above the street level. What is the initial speed of the ball? What is the angle above the horizontal at which it was thrown?

Answer 1

If it takes 5 seconds to cover that distance you will have to throw it just about straight up so it will stay up five seconds !

u = constant horizontal speed = 10m/5s
= 2 meters/second

in 5 seconds it goes up 3 meters

3 = Vi t - 4.9 t^2
3 = Vi(5) = 4.9(25)
3 = 5 Vi - 122.5
Vi = 25.1 m/s
so
speed = sqrt (25.1^2 + 2^2)
= 25.2 m/s
tan T = Vi/u = 25.2/2
T = 85.5 degrees