a ball rolls down a roof that makes an angle of 30 to horizontal.it rolls off the egde with a speed of 5.00m/s.distance to the ground from that point is 7.00m

a)how long is the ball in the air
b)how far from the base of the house does it land?
c)what is its speed just before landing?

quadratic equation is solved incorrectly 4.9•t^2 +2.5•t-7 =0

t IS NOT = 1.05 s , it is 0,967047 seconds.
then put this time to solve other stuff

v(ox) = v•cosα =5•cos30 = 4.33 m/s,

v(x) = v(ox) = 4.33 m/s,
v(oy) = v•sinα =5•sin30 = 2.5m/s.
h =v(oy) •t+g•t^2/2,
4.9•t^2 +2.5•t-7 =0
t = 1.05 s.
L =v(ox) •t = 4.33•1.05 = 4.55 m
v(y) = v(oy) + g•t = 2.5 +9.8•1.05 = 12.8 m/s,
v=sqrt{v(x)^2 + v(y)^2} = 13.5 m/s.

To answer these questions, we can use the equations of motion in physics. Let's break it down step by step:

a) How long is the ball in the air?

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

s = ut + (1/2)gt^2

where:
s = vertical distance covered (7.00 m)
u = initial vertical velocity (0 m/s, as the ball starts from rest)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time

Rearranging the equation to solve for time (t), we get:

7.00 = 0.5(-9.8)t^2

Simplifying further:

(-9.8)t^2 = 7.00 × 2

t^2 = (7.00 × 2) / (-9.8)

t^2 = 1.4286

Taking the square root of both sides, we find:

t ≈ √1.4286

t ≈ 1.1967 seconds

Therefore, the ball is in the air for approximately 1.1967 seconds.

b) How far from the base of the house does it land?

To find the horizontal distance the ball travels, we can use the equation of motion for horizontal motion:

s = ut

where:
s = horizontal distance covered
u = initial horizontal velocity (velocity remains constant throughout horizontal motion)
t = time

The initial horizontal velocity, u, can be found using trigonometry. The ball rolls off the edge of the roof with a speed of 5.00 m/s, and the angle of the roof with respect to the horizontal is 30 degrees.

Using the trigonometric relationship, we can find the horizontal component of velocity (u_x):

u_x = u * cosθ

where:
u_x = horizontal component of velocity
u = initial speed (5.00 m/s)
θ = angle with respect to horizontal (30 degrees)

Calculating for u_x:

u_x = 5.00 * cos(30°)

u_x = 5.00 * √(3/2)

u_x ≈ 4.33 m/s

Now, let's substitute the values we have into the equation:

s = (4.33 m/s) * (1.1967 s)

s ≈ 5.18 m

Therefore, the ball lands approximately 5.18 meters from the base of the house.

c) What is its speed just before landing?

To find the speed of the ball just before it lands, we can use the equation of motion for vertical motion:

v = u + gt

where:
v = final vertical velocity (0 m/s, as the ball comes to rest when it lands)
u = initial vertical velocity (0 m/s, as the ball starts from rest)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time

Rearranging the equation to solve for u, we get:

v = u + gt

0 = 0 + (-9.8) × 1.1967

u = 9.8 × 1.1967

u ≈ 11.73 m/s

Therefore, the speed of the ball just before it lands is approximately 11.73 m/s.

First of all I guess Elena has never taking a high school physics am I right guys XD #tryagain