vertical problem:
2.5 seconds up and 2.5 seconds down
v = Vi - 9.81 t
at top v = 0
so
Vi = 9.81*2.5 = 24.5 m/s up
h = Vi t - 4.9 t^2
= 24.5(2.5) -4.9(6.25)
= 61.3 - 30.6
= 30.6 meters high at top
horizontal problem
49 = u (5)
so u = 9.8 m/s
tan theta = Vi/u = 24.5/9.8
theta = 68.2 degrees
a ball that was projected up at an angle of θ degrees with the horizontal has a flight time of 5.0s and covers a range of 49.0 m. find (a) the time it took to reach maximum height, (b) the horizontal and vertical components of its initial velocity, (c) its initial velocity,(d) the angle θ at which it was projected, and (e) the maximum height it reached.
2 answers
the horizontal and vertical component of its initial velocity of 5.0s and 49.0m