Asked by Ayishatu
A stone propelled from a catapult with a speed of 100m/s and attains a height of 50 m .What would be the horizontal distance traveled
Answers
Answered by
henry2,
Y^2 = Yo^2 + 2g*h = 0 m/s.
Yo^2 + (-19.6)50 = 0,
Yo = 31.3 m/s. = Ver. component of initial velocity.
Yo = Vo*sinA = 31.3.
100*sinA = 31.3,
sinA = 0.313,
A = 18.2o.
Xo = Vo*CosA = 100*Cos18.2 = 95 m/s. = Hor. component of Vo.
Y = Yo + g*t = 0.
31.3 + (-9.8)t = 0,
t = 3.19 s. = Rise time.
d = Xo * t = 95 * 3.19 = meters.
Yo^2 + (-19.6)50 = 0,
Yo = 31.3 m/s. = Ver. component of initial velocity.
Yo = Vo*sinA = 31.3.
100*sinA = 31.3,
sinA = 0.313,
A = 18.2o.
Xo = Vo*CosA = 100*Cos18.2 = 95 m/s. = Hor. component of Vo.
Y = Yo + g*t = 0.
31.3 + (-9.8)t = 0,
t = 3.19 s. = Rise time.
d = Xo * t = 95 * 3.19 = meters.