A ball is tossed from an upper-story window of a building. The ball is given an

Question

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 37.0° below the horizontal. The ball hits the ground 3.00sec later. Set your x-y axis:a) Write the position of the object (x, and y) as a function of time b) How far horizontally from the base of the building does the ball strike the ground? c) Find the height from which the ball was thrown. d) What is the magnitude and the direction of the velocity of the ball just before it hits the ground?

Explanation / Answer

Vi = 8.0 @ 37.0o

t = 3.0 s
Initial Horizontal Velcoity, Vx = 8.0 * cos(37) = 6.39 m/s
Initial Vertical Velocity, Vy = 8.0 * sin(37) = 4.81 m/s

Horizontal acceleration, = 0
Vertical acceleration, = 9.8 m/s^2
(a)
x = Vx*t
y = Vy*t + 1/2*ay*t^2
(b)
How far horizontally from the base of the building does the ball strike the ground?
Distance = Velocity * time
Sx = 6.39 * 3.0 m
Sx = 19.17 m

(c)
Let the height of the building = h
Now,
h = u*t + 1/2 *at^2
h = 4.81 * 3.0 + 1/2 * 9.8 * 3.0^2
h = 58.5 m
height from which ball was thrown, h = 58.5 m

(d)
Just before it hits the ground.
Vertical Velocity,
Vy = Vi + g*t
Vy = 4.81 + 9.8*3.0
Vy = 34.2 m/s
Horizontal Velocity, Vx = 6.39 m/s

|Vnet| = sqrt(34.2^2 + 6.39^2)
|Vnet| = 34.8 m/s
Direction = tan^-1(34.2/6.39)
Direction = 79.4 o Below Horizontal.

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