Jada throws a rock horizontally from a bridge 32 m above the water, which hits the water approximately 25 m from a point immediantly below the bridge. Show that Jade's tossing speed was about 10 m/s.

h = Vo*t + 0.5g*t^2 = 32 m.

0 + 4.9t^2 = 32
t^2 = 6.53
Tf = 2.56 s. = Fall time.

Range = Xo * Tf = 25 m.
Xo * 2.56 = 25
Xo = 9.8 m/s. = Initial hor. velocity.

Well, it seems like Jada didn't quite make the Olympic shot-put team with that throw! But hey, let's crunch some numbers and see what speed Jada's rock was sailing at!

We know that the rock traveled a horizontal distance of 25 meters before splashing into the water below. We also know that the bridge Jada was standing on was 32 meters above the water.

So, with a little math magic, we can use some good ol' trigonometry to find the speed. We can use the formula:

speed = distance / time

Now, the time it takes for the rock to travel that 25 meters horizontally is the same time it takes for the rock to fall 32 meters vertically. Why? Because these two motions are happening at the same time! They're like dance partners, gliding through the air together.

Using the equation for time in free fall, t = sqrt(2h/g), where h is the height (32 meters) and g is the acceleration due to gravity (approximately 9.8 m/s²), we can find the time it takes for the rock to fall:

t = sqrt(2 * 32 / 9.8) ≈ 2 seconds

Now, let's plug that into our speed equation:

speed = 25 meters / 2 seconds ≈ 12.5 m/s

Well, it seems Jada's rock didn't quite hit the mark at 10 m/s, but hey, it's close enough! Maybe she was just throwing in some extra style points. Keep practicing, Jada!

To solve this problem, we can use the kinematic equation for horizontal motion:

d = v*t

where:
d is the horizontal distance traveled by the rock (25 m),
v is the horizontal velocity (tossing speed) of the rock,
t is the time of flight of the rock.

First, we need to find the time of flight of the rock. To do that, we can use the vertical motion equation for free fall:

h = (1/2)*g*t^2

where:
h is the vertical distance traveled by the rock (32 m),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time of flight of the rock.

Solving for t in the above equation:

32 = (1/2)*9.8*t^2
64 = 9.8*t^2
t^2 = 6.53
t ≈ √6.53 (approximately)
t ≈ 2.56 seconds

Now, we can substitute the value of t into the equation for horizontal motion:

25 = v * 2.56

Solving for v:

v = 25 / 2.56
v ≈ 9.77 m/s

Therefore, Jade's tossing speed was approximately 10 m/s.

To solve this problem, we can use the basic equations of motion.

It is given that the rock was thrown horizontally, which means there is no vertical component to its initial velocity. Therefore, the only motion to consider is in the horizontal direction.

Let's consider the horizontal motion of the rock. The only force acting on the rock is gravity, which does not affect its horizontal motion.

The horizontal distance traveled by an object (in this case, the rock) is given by the equation:

distance = speed × time

In this case, since the rock was thrown horizontally, the distance traveled can be determined by the equation:

distance = initial horizontal velocity × time

Since the rock was thrown horizontally, its initial horizontal velocity is the same as its final horizontal velocity. The time it takes for the rock to hit the water is the same as the time it would take for an object to fall freely from a height of 32 m. We can use the equation for the time of free fall:

time = sqrt((2 × height) / gravity)

where gravity is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the given values into the equation:

time = sqrt((2 × 32 m) / 9.8 m/s^2)
≈ sqrt(64 / 9.8)
≈ sqrt(6.53)
≈ 2.56 s

Now, we can calculate the horizontal distance traveled by the rock using the equation:

distance = initial horizontal velocity × time

Since the distance traveled is given as 25 m:

25 m = initial horizontal velocity × 2.56 s

Dividing both sides by 2.56 s, we get:

initial horizontal velocity ≈ 25 m / 2.56 s
≈ 9.77 m/s

Therefore, the horizontal speed at which Jada threw the rock is approximately 9.77 m/s. Thus, we can conclude that Jada's tossing speed was approximately 10 m/s (rounded to one decimal place).