A ball is shot from the ground into the air. At a height of 9.4 m, its velocity

Question

A ball is shot from the ground into the air. At a height of 9.4 m, its velocity is v Overscript right-arrow EndScripts equals 7.3 i Overscript EndScripts plus 7.4 j Overscript EndScripts m divided by s, with i Overscript EndScripts horizontal and j Overscript EndScripts upward. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball travel? What are the (c) magnitude and (d) angle (below the horizontal; give as negative) of the ball's velocity just before it hits the ground?

Explanation / Answer

here,

velocity at height, h is v = (7.3 i + 7.4 j) m/s

Maximum height
from third eqn of motion
h = vy^2/2*g
h = 7.4^2/(2*9.81)
h = 2.791 m

Therefore, net height, Ho = 2.791 + 9.4 = 12.191 m

time of flight
t = 2*v/g = 2*sqrt(7.3^2 + 7.4^2 )/9.81
t = 2.119 s

Horizontal Distance, x
x = vx*t
x = 7.3 * 2.119
x = 15.469 m

At the time of impact,
Vx = 7.3 m/s ( always)

Vy = - sqrt(2*g*hmax)
Vy = - sqrt(2*9.81*12.191)
Vy = - 15.466 m/s

Magnitude of Velocity, V = sqrt(7.3^2 + (-15.466)^2)
V = 17.102 m/s

Angle , A = arctan(y/x) = arctan(15.466/7.3)
A = 64.733 degrees below horizontal

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