Asked by Anonymous
A blue ball is dropped from a height of 3.8 m above a classroom floor while a red ball held at the same height is thrown horizontally with a speed of 5 m/sec. Find
The initial velocity and acceleration of each ball (the positive x-direction is the direction the ball is thrown and the positive y-direction is upward.)
The time required for each ball to hit the floor.
The horizontal position of each ball when it lands.
The final velocity and acceleration of each ball.
The initial velocity and acceleration of each ball (the positive x-direction is the direction the ball is thrown and the positive y-direction is upward.)
The time required for each ball to hit the floor.
The horizontal position of each ball when it lands.
The final velocity and acceleration of each ball.
Answers
Answered by
Damon
The vertical problem is the same for both
Vi = 0
a = -9.8 until it stops
Hi = 3.8
v = 0 + a t = -9.8 t
h = 3.8 + 0 t - 4.9 t^2
when h = 0, the floor
4.9 t^2 = 3.8
t = .88 seconds in the air for both
v at floor = -9.8 (.88) = - 8.62 m/s
first ball straight down, x = 0
second ball u = 5 so x = 5(.88)
first ball falls straight down -8.62
second ball -8.62 down and 5 hor
so speed = sqrt (8.62^2+5^2)
angle down from horizontal
tan A = 8.62/5
Vi = 0
a = -9.8 until it stops
Hi = 3.8
v = 0 + a t = -9.8 t
h = 3.8 + 0 t - 4.9 t^2
when h = 0, the floor
4.9 t^2 = 3.8
t = .88 seconds in the air for both
v at floor = -9.8 (.88) = - 8.62 m/s
first ball straight down, x = 0
second ball u = 5 so x = 5(.88)
first ball falls straight down -8.62
second ball -8.62 down and 5 hor
so speed = sqrt (8.62^2+5^2)
angle down from horizontal
tan A = 8.62/5
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