A ball is thrown horizontally off a 35 m tall building. It hits the ground 80 m from the bottom of the tower.

a. what is the balls time in the air
[i got 2.67]
b. what is its initial velocity?
[i got 0]
c. What is the magnitude and direction of its velocity as it strikes the ground?

Could someone please check my answers and explain how to do question c??

b is wrong. should be a vector in the x dircection. will add more later

Well it is thrown horizontally so its initial velocity in the vertical direction is zero. That means it just plain falls while continuing at the original horizontal velocity. So:

Time to fall 35 m
0 = 35 + 0 - 4.9 t^2
t^2 = 35/4.9
t = 2.67 seconds in the air

t went 80 meters in 2.67 seconds so initial and final and in between horizontal velocity is 80/2.67 = 30 m/s

Well we know the horizontal velocity is zero during this whole experience so we need the vertical velocity when it falls for 2.67 seconds
v = a t = - 9.8 * 2.67 = - 26.2 m/s
so when it crashes it is still going 30 ahead and is going 26.2 down
magnitude = sqrt(26.2^2 +30^2) = 39.8 m/s
tangent of angle down from horizontal = 26.2/30
so angle = 41.1 degrees

Well it is thrown horizontally so its initial velocity in the vertical direction is zero. That means it just plain falls while continuing at the original horizontal velocity. So:

Time to fall 35 m
0 = 35 + 0 - 4.9 t^2
t^2 = 35/4.9
t = 2.67 seconds in the air

It went 80 meters in 2.67 seconds so initial and final and in between horizontal velocity is 80/2.67 = 30 m/s

Well we know the horizontal velocity is 30 m/s during this whole experience so we need the vertical velocity when it falls for 2.67 seconds
v = a t = - 9.8 * 2.67 = - 26.2 m/s
so when it crashes it is still going 30 ahead and is going 26.2 down
magnitude = sqrt(26.2^2 +30^2) = 39.8 m/s
tangent of angle down from horizontal = 26.2/30
so angle = 41.1 degrees

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To calculate the time it takes for the ball to reach the ground, we can use the equation of motion for vertical motion:

h = (1/2)gt²

Where:
h is the height of the building (35m),
g is the acceleration due to gravity (-9.8 m/s²), and
t is the time in seconds.

Rearranging the equation, we get:

t = sqrt(2h/g)

Substituting the values:

t = sqrt(2 * 35 / 9.8)
t ≈ 3.19 seconds (rounded to two decimal places)

Therefore, the ball's time in the air is approximately 3.19 seconds.

To calculate the initial velocity (u) of the ball, we can use the equation of motion for horizontal motion:

s = ut

Where:
s is the horizontal distance traveled (80m),
u is the initial horizontal velocity, and
t is the time in seconds.

Rearranging the equation, we get:

u = s / t

Substituting the values:

u = 80 / 3.19
u ≈ 25.08 m/s (rounded to two decimal places)

Therefore, the ball's initial velocity is approximately 25.08 m/s.

For the final part, let's consider the horizontal and vertical components of the ball's velocity when it hits the ground. Since there is no vertical acceleration (neglecting air resistance), the vertical component of the velocity remains constant throughout the motion.

The magnitude (or speed) of the ball's velocity can be found using the Pythagorean theorem:

v = sqrt(v(horizontal)² + v(vertical)²)

We know the initial velocity in the horizontal direction is 25.08 m/s, and there is no vertical acceleration, so the vertical component of the velocity remains constant. The magnitude of the velocity at the time of impact can be calculated as:

v = sqrt(25.08² + 0²)
v ≈ 25.08 m/s (rounded to two decimal places)

Therefore, the magnitude of the ball's velocity as it strikes the ground is approximately 25.08 m/s.

Regarding the direction of the velocity, since the ball is thrown horizontally, the initial velocity in the vertical direction is zero. Therefore, the direction of the velocity as it strikes the ground is horizontal.