baseball is hit with a speed of 30.0 m/s at an angle of 50 It lands on the flat roof of a 10m tall nearby building, If the ball was hit when it was 1.5 m above the ground, what horizontal distance does it travel before it lands on the building?

V 0x = 30*cos 50 = 19.28 m/s

V 0y = 30*sine 50 = 22.98

Y = 14-1.5 =12.5
Y = V 0y*t -0.5*g*t^2
12.5 = 22.98*t -4.9*t^2
Solve the quadratic equation for t. t = 4.06 seconds

X = v 0x*t
X = 19.28*4.06 = 78.27 meters

To find the horizontal distance the baseball travels before it lands on the building, we need to break down the problem using the projectile motion equations.

First, let's find the time it takes for the baseball to hit the building. We can use the vertical motion equation for the height:

y = y0 + v0yt - (1/2)gt^2

Where:
- y is the final height (10 m),
- y0 is the initial height (1.5 m),
- v0y is the vertical component of the initial velocity (v0 * sinθ),
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time.

Using the given values, we can rearrange the equation to solve for t:

10 = 1.5 + (30 * sin(50))t - (1/2)(9.8)t^2

Multiply the equation by 2 to eliminate the fraction:

20 = 3 + (60 * sin(50))t - 9.8t^2

Rearrange to form a quadratic equation:

9.8t^2 - (60 * sin(50))t + 17 = 0

To solve this equation, we can either use the quadratic formula or factor it. Let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Given:
a = 9.8, b = -(60 * sin(50)), c = 17

Substituting these values into the formula:

t = (-(60 * sin(50)) ± √((60 * sin(50))^2 - 4 * 9.8 * 17)) / (2 * 9.8)

Calculating this expression will give us two possible values for t. We'll use the positive value, as negative time doesn't make sense in this context.

Next, we need to find the horizontal distance the baseball travels during this time. We can use the horizontal motion equation:

x = x0 + v0xt

Where:
- x is the horizontal distance,
- x0 is the initial horizontal position (0 m since the ball was hit at ground level),
- v0x is the horizontal component of the initial velocity (v0 * cosθ),
- t is the time.

Calculate v0x using the given values:

v0x = 30 * cos(50)

Substituting the known values into the equation:

x = 0 + (30 * cos(50))t

Finally, substitute the value of t previously calculated to find the horizontal distance:

x = (30 * cos(50)) * t

To find the horizontal distance the baseball travels before it lands on the building, we can use the range formula for projectile motion. The range formula is given by:

R = (V₀² * sin(2θ)) / g

where:
R is the horizontal range or distance,
V₀ is the initial velocity of the baseball,
θ is the launch angle, and
g is the acceleration due to gravity.

In this case, the initial velocity V₀ is given as 30.0 m/s and the launch angle θ is given as 50°. The acceleration due to gravity g can be taken as 9.8 m/s².

First, we need to find the time taken by the baseball to reach the height of 10m on the building. We can use the vertical motion equation:

h = V₀ * sin(θ) * t - (1/2) * g * t²

where:
h is the height or displacement,
t is the time taken, and
V₀ and θ are as defined above.

In this case, h is 10m, V₀ is 30.0 m/s, θ is 50°, and g is 9.8 m/s².

Plugging in these values, we get:

10 = (30 * sin(50) * t) - (1/2) * (9.8) * t²

Simplifying this equation, we get a quadratic equation which can be solved to find the time t.

Once we have the time t, we can substitute it back into the range formula to find the horizontal distance R traveled by the baseball before it lands on the building.

Let's calculate the values step by step:

1. Calculate the time taken to reach a height of 10m on the building using the quadratic equation.
2. Substitute the value of time t in the range formula to find the horizontal distance R.

By following these steps, you should be able to find the horizontal distance the baseball travels before it lands on the building.

The ball travels in the vertical direction with initial vertical component of 30 sin 50 [22.98 m/s^2] upward till its velocity component in vertical direction becomes let us say V. The final vertical component after travelling [14 - 1.5] = 12.5 m is given by

[30 sin 50]^2 - V^2 = 2x9.8x12.5 or

V^2 = 528.14 - 245 = 283.14 or V = 16.826 m/s

Time required to reduce the vertical component from 22.98 to 16.82 would be 6.16/9.8 s = 0.628 s

The horizontal distance travelled by the ball in this time is

30 cos 50 x 0.628 = 12.11 m