A ball is thrown from the top of a building upward at an angle of 45 to the hori

Question

A ball is thrown from the top of a building
upward at an angle of 45 to the horizontal
and with an initial speed of 22 m/s. The ball
is thrown at a height of 51 m above the ground
and hits the ground 80.6274 m from the base
of the building.What is the speed of the ball just before it
strikes the ground? The acceleration due to
gravity is 9.8 m/s2 .
Answer in units of m/s

Question 2

A particle moving at a velocity of 6.1 m/s in
the positive x direction is given an accelera-
tion of 5.3 m/s2 in the positive y direction for
1.6 s.
What is the final speed of the particle?
Answer in units of m/s

Explanation / Answer

The y (vertical) position of the ball can be modeled as:

y = 51 + 22 sin(45o) t - (1/2)gt2

The x (horizontal) position of the ball can be modeled as:

x = 0 + 22 cos(45o) t - 0t2

x = 22 cos(45o) t

When it finally lands, x is the distance it went from the base of the building so we can solve for t:

80.6274 = 22 cos(45o) t

t = 5.183 s

The y component of the velocity of the ball is given by the derivative of the position so:

vy = 22 sin(45o) - gt

Likewise:

vx = 22 cos(45o)

So when it hits the ground:

vy = 22 sin(45o) - g(5.183) = -35.237 m/s

vx = 22 cos(45o) = 15.556 m/s

The speed is the magnitude of the velocity: ((-35.237)2 + (15.556)2) = 38.518 m/s

Question 2:

vx = 6.1

vy = (5.3)t for t from 0 to 1.6

vy = (5.3)(1.6) = 8.48 for t > 1.6

This is because the acceleration is only acting on the object for 1.6 seconds, so the final and permanent velocity in the y-direction is given by the initial velocity (0) plus the acceleration applied over the amount of time it is applied.

Speed is again the magnitude of the velocity: ((6.1)2+(8.48)2) = 10.446 m/s

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