s = speed
45.9 = u T
where
u = s cos 40
so
T = 45.9 /(s cos 40) where T is time aloft
rise time is half T or t = T/2
v = Vi - g t
Vi = s sin 40
at top v = 0
s sin 40 = 9.81 t = 9.81[45.9/(2s cos 40]
or
2 s^2 sin 40 cos 40 = 9.81*45.9
remember trig now
A football quarterback shows off his skill by throwing a pass 45.90 m downfield and into a bucket. The quarterback consistently launches the ball at 40.00 ∘ above horizontal, and the bucket is placed at the same level from which the ball is thrown.
What initial speed is needed so that the ball lands in the bucket?
2 answers
Similarly, when you solve for m = 46, you get:
2 s^2 sin 40 cos 40 = 9.81*46
s = 21.40 m/s
You're welcome.
2 s^2 sin 40 cos 40 = 9.81*46
s = 21.40 m/s
You're welcome.