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An object is shot from ground level at an angle of 50 from the horizontal with an initial velocity of 30m/s?
An object is shot from ground level at an angle of 50 from the horizontal with an initial velocity of 30m/s
a)Find the time for the object to reach maximum height
b)Calculate the maximum height
c)Find the range of the object
d)Find the horizontal and vertical components of the velocity of the object when it reaches the ground.
5 points Thanks!
1 Answer
- QuarkLv 610 years agoFavorite Answer
(a) t = vi sin α/g = 30 * sin 50/9.8 = 2.35 s
(b) hmax = vi^2 * sin^2 α/2g = 30^2 * ( sin 50)^2/2 * 9.8 = 26.95 m
(c) R = vi^2 * sin 2α/g = 30^2 * sin 100/9.8 = 90.44 m
(d) vx = vi cos α = 30 * c0s 50 = 19.28 m/s
vy = vi sin α - gt
= 30 * sin 50 - 9.8 * ( 2 * 2.35)
= -23.08 m/s
