A ball is thrown at an angle of 45 degrees to the ground. If the ball lands 90m away, what is the initial speed of the ball? I know you use x(t)=(v*cos(alpha))t and y(t)=(v*sin(alpha))t-g/2t^2 but I'm still really confused about what to do. Please show all work. Thank you.

Well, I can definitely help you with that! Let's break it down step by step.

We know that the ball is thrown at an angle of 45 degrees to the ground. This means that the initial velocity of the ball can be split into horizontal and vertical components.

The horizontal component of the velocity (Vx) remains constant throughout the entire trajectory as there is no horizontal acceleration. The vertical component of the velocity (Vy) changes due to the effect of gravity.

Now, let's consider the given equation for the x-coordinate at time t:

x(t) = (V * cos(alpha)) * t

Here, alpha represents the angle of projection (45 degrees), and we need to find V, the initial speed of the ball.

Since the ball lands 90m away, we can substitute these values into the equation:

90 = (V * cos(45)) * t

Next, let's look at the equation for the y-coordinate at time t:

y(t) = (V * sin(alpha)) * t - (1/2) * g * t^2

Here, g represents the acceleration due to gravity. Substituting the given values, we have:

0 = (V * sin(45)) * t - (1/2) * g * t^2

Now, since the time of flight (t) is the same for both x(t) and y(t), we can equate them:

(V * cos(45)) * t = (V * sin(45)) * t - (1/2) * g * t^2

Simplifying this equation further, we get:

2 * (V * cos(45)) * t = (V * sin(45)) * t

Dividing by t, we have:

2 * (V * cos(45)) = V * sin(45)

Now, let's break down the trigonometric functions:

2 * (V * (√2/2)) = V * (√2/2)

Simplifying further:

V = V * (√2/2)

At this point, we can cancel out V from both sides of the equation:

1 = (√2/2)

That's impossible! I must have made a mistake in my calculations. My apologies for the confusion. I'll have to recalculate and get back to you.

To solve this problem, we can use the given information and the equations of motion for projectile motion.

Given:
Angle of projection (alpha) = 45 degrees
Distance traveled in the x direction = 90 m
Gravity (g) = 9.8 m/s^2

We can start by analyzing the motion in the x and y directions separately.

Motion in the x-direction:
The horizontal motion is unaffected by gravity, so we can use the equation x(t) = (v * cos(alpha)) * t.

Since the ball lands at a distance of 90 m, we can write:
90 = (v * cos(45)) * t

Motion in the y-direction:
The vertical motion is affected by gravity, so we can use the equation y(t) = (v * sin(alpha)) * t - (g/2) * t^2.

Since the ball lands on the ground, the vertical displacement is 0, so we can write:
0 = (v * sin(45)) * t - (9.8/2) * t^2

Now we have two equations with two unknowns: v (initial speed) and t (time of flight). We can solve these equations simultaneously.

From the equation 90 = (v * cos(45)) * t, we can simplify:
t = 90 / (v * cos(45))

Substituting this value of t into the equation 0 = (v * sin(45)) * t - (9.8/2) * t^2, we get:
0 = (v * sin(45)) * (90 / (v * cos(45))) - (9.8/2) * (90 / (v * cos(45)))^2

Simplifying further:
0 = sin(45) * 90 - (9.8/2) * (90^2) / (v * cos(45))^2

Now, we can solve this equation for v. Let's calculate:

0 = sin(45) * 90 - (9.8/2) * (90^2) / (v * cos(45))^2

0 = 63.64 - (9.8/2) * (90^2) / (v^2)

0 = (2 * 63.64 * v^2) - (9.8/2) * (90^2)

Now, we can solve this quadratic equation for v. The solution will give us the initial speed of the ball.

(2 * 63.64 * v^2) - (9.8/2) * (90^2) = 0

Simplifying the equation further:
127.28 * v^2 - 3969 = 0

Now, we can solve this quadratic equation to find v:

v^2 = 3969 / 127.28

v^2 = 31.2

Taking the square root of both sides:
v ≈ √31.2

v ≈ 5.59 m/s

Therefore, the initial speed of the ball is approximately 5.59 m/s.

To solve this problem, we need to analyze the motion of the ball in two separate directions: horizontal (x-axis) and vertical (y-axis). We can then use the given information to find the initial speed (v).

Let's begin by breaking down the projectile motion equations you mentioned:

Horizontal motion equation:
x(t) = (v * cos(alpha)) * t

Vertical motion equation:
y(t) = (v * sin(alpha)) * t - (1/2) * g * t^2

where:
x(t) = horizontal displacement at time t
y(t) = vertical displacement at time t
v = initial speed of the ball
alpha = angle of projection (45 degrees in this case)
g = gravitational acceleration (approximately 9.8 m/s^2)

Since we are given that the ball lands 90m away (horizontal displacement, x(t) = 90 m), we can substitute this value into the horizontal motion equation:

90 = (v * cos(45)) * t

Simplifying this equation, we have:

90 = (v * √(2) / 2) * t [since cos(45) = √(2) / 2]

Dividing both sides of the equation by (v * √(2) / 2), we get:

180 / (v * √(2)) = t

Now, let's consider the vertical motion equation. Since we are interested in the time it takes for the ball to land, we can set y(t) = 0 and solve for t:

0 = (v * sin(45)) * t - (1/2) * g * t^2

0 = (v * √(2) / 2) * t - (1/2) * (9.8) * t^2 [using sin(45) = √(2) / 2 and g = 9.8 m/s^2]

This equation can be simplified to:

0 = (v * √(2) / 2) * t - 4.9 * t^2

Now we have a system of equations with two variables (v and t). We can solve these equations simultaneously to find the values of v and t.

Substituting the value of t from the first equation into the second equation, we get:

0 = (v * √(2) / 2) * (180 / (v * √(2))) - 4.9 * (180 / (v * √(2)))^2

0 = 90/v - 4.9 * (90/v)^2

Dividing through by v^2, we have:

0 = 90/v^3 - 4.9 * (90/v)^2

Now, you can solve this cubic equation numerically using a graphing calculator, computer software, or online equation solver. The resulting value of v will give you the initial speed of the ball.

The question is missing information.

By matching different initial velocities of the ball, it can land at 90m away at virtually any angle α.

I will solve for the initial velocity in terms of the angle α, but also give a solution for the MINIMUM initial velocity for the ball to land 90m away.

The key is to recognize that
x(t)=90m=D, and
y(t)=0
(i.e. the ball lands D=90m away).

Using the second formula, we have
y(t)=(v*sin(alpha))t-(g/2)t^2, or
(v*sin(alpha))t=(g/2)t^2

Solve for t after cancelling t on each side:
t=2vsin(α)/g

Substitute in the first equation to solve for v(cos(α):
D=vcos(α)*(2vsin(α)/g
Substitute
sin(2α) = 2sin(α)cos(&alpha)
D=v²sin(2α)/g
or
v²=Dg/sin(2α)

(The following is the only place where calculus comes in)

The minimum velocity that can give the distance D is obtained when the ball is thrown at 45° with the horizontal, in which case α=45°, or when 2α=90°.

In this case the (minimum) initial velocity is
v=sqrt(90*9.81)=29.71m/s