1) A T-shirt cannon at a basketball game launches shirts to the upper level at 55 m/s with an angle of 52 degrees. Find the height (dy) of the seats if it takes 4.5 seconds to land in the seats.

2) A soccer ball is kicked on a flat field by the goalie at an initial velocity of 22 m/s and an angle of 30 degrees. How far downfield does it land ?

1. Vo = 55m/s[52o].

Xo = 55*Cos52 = 33.9 m/s.
Yo = 55*sin52 = 43.3 m/s.

h = Yo*t + 0.5g*t^2 =
g = -9.8 m/s^2.
h = ?.

2. Range = Vo^2*sin(2A)/g.
Vo = 22 m/s.
A = 30o.
g = 9.8 m/s^2.

1) To find the height (dy) of the seats, we can use the equation of motion:

dy = vy * t - (1/2) * g * t^2

where:
dy is the height of the seats,
vy is the vertical component of the initial velocity,
t is the time taken to land, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).

First, let's find the vertical component of the initial velocity (vy):
vy = velocity * sin(angle)
vy = 55 m/s * sin(52 degrees)
vy ≈ 55 m/s * 0.788
vy ≈ 43.34 m/s

Now, substitute the values into the equation:
dy = (43.34 m/s) * (4.5 s) - (1/2) * (9.8 m/s^2) * (4.5 s)^2
dy ≈ 195.03 m - 99.44 m
dy ≈ 95.59 m

Therefore, the height (dy) of the seats is approximately 95.59 meters.

2) To find how far downfield the soccer ball lands, we can use the range formula:

Range = velocity * cos(angle) * time

where:
Range is the horizontal distance covered,
velocity is the initial velocity,
angle is the launch angle, and
time is the time taken to land.

Let's plug in the given values:
Range = 22 m/s * cos(30 degrees) * time

We know that the time will be the same as the time taken to land, as mentioned in the question.

Since no time value is provided, we'll need more information to calculate the distance.

To solve both of these questions, we can use the equations of motion for projectile motion. We can break down the initial velocity into horizontal and vertical components, and use the equations of motion to find the desired quantities.

1) Let's solve the first question step by step.

Given:
Initial speed (v0) = 55 m/s
Angle (θ) = 52 degrees
Time of flight (t) = 4.5 seconds

First, let's find the initial vertical velocity (v0y) and the initial horizontal velocity (v0x):

v0y = v0 * sin(θ)
v0x = v0 * cos(θ)

Next, we can find the time at which the object reaches maximum height (t_max). At maximum height, the vertical velocity becomes zero:

0 = v0y - g * t_max
t_max = v0y / g

Using this, we can also find the maximum height (h_max):

h_max = v0y * t_max - (1/2) * g * t_max^2

Now, we need to find the time it takes for the object to reach the seats. Since the total time of flight is given, we can subtract the time to reach maximum height to find this:

t_seats = t - t_max

Finally, we can calculate the height of the seats:

dy = v0y * t_seats - (1/2) * g * t_seats^2

Using the values provided, you can substitute them into the equations and calculate the height (dy) of the seats.

2) Let's solve the second question step by step.

Given:
Initial speed (v0) = 22 m/s
Angle (θ) = 30 degrees

First, let's find the initial vertical velocity (v0y) and the initial horizontal velocity (v0x):

v0y = v0 * sin(θ)
v0x = v0 * cos(θ)

Since the ground is flat, the time of flight (t) in projectile motion is symmetrical. So, the time taken to reach the maximum height is the same as the total time taken to reach the ground.

t = 2 * t_max

Now, we can find the range (R), which is the distance the ball travels horizontally:

R = v0x * t

Substituting the given values, we can calculate the distance downfield that the soccer ball lands.

By using these equations and substituting the values given in each problem, you will be able to find the desired answers.