HELP A ball is thrown from the ground into the air. At a height of 9.1 m, the ve
ID: 2126178 • Letter: H
Question
HELP
A ball is thrown from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = 7.1i + 6.8j in meters per second (x axis horizontal, y axis vertical and up).
A ball is thrown from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = 7.1i + 6.8j in meters per second (x axis horizontal, y axis vertical and up). To what maximum height will the ball rise? m What will be the total horizontal distance traveled by the ball? m What is the magnitude of the ball's velocity the instant before it hits the gound? m/s What is the direction of the ball's velocity the instant before it hits the ground?Explanation / Answer
Vertically:
6.8^2 = u2^2 - 2g * 9.1
u2^2 = 6.8^2 + 2g * 9.1
u2 = 15.0 m/s.
Horizontally there is no change in velocity:
u1 = 7.2 m/s ...(1)
(a)
At the maximum height h m, there is no vertical velocity.
0 = u2^2 - 2gh ...(2)
h = u2^2 / 2g
h = 11.5m.
(b)
Let t be the total time of flight,
0 = u2*t - gt^2 / 2
= t(u2 - gt / 2)
t = 2u2 / g = 30/g
x = u1*t = 7.2t = 7.2 * 30 / g
x = 22.0m.
Let the ball's velocity just before landing be (v1 i + v2 j).
(c)
Vertically:
v2^2 = 2g * 9.1
Comparing this equation with (2), v2 = - u2 = - 15.0j m/s.
Horizontally, from (1):
v1 = 7.2i m/s.
v = sqrt(v1^2 + v2^2) = sqrt(15^2 + 7.2^2)
v = 16.6 m/s.
(d)
If t is the angle above the horizontal,
tan(a) = v2 / v1 = - 15.0 / 7.2
a= - 64.4 deg, indicating an angle below the horizontal.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.