A home run is hit such a way that the baseball

just clears a wall 27 m high located 147 m
from home plate. The ball is hit at an angle
of 37◦ to the horizontal, and air resistance is
negligible. Assume the ball is hit at a height
of 1 m above the ground.
The acceleration of gravity is 9.8 m/s2 .
What is the initial speed of the ball?
Answer in units of m/s.
---------------------------------

I tried using these equations
xf=xi+vxit+.5g^2
tan theta= vy/vx => vy= vx*tan theta
yf=vyt+.5gt^2

147= 0+ vxi(4.30407) + 4.9(4.30407)^2=13.0638 wrong
I got t= 4.30407 by pluging in vy= vx*tan theta into yf=vyt+.5gt^2
1. The problem statement, all variables and given/known data

147= 0+ vxi(4.30407) + 4.9(4.30407)^2

well, I have issues with this. Why is gravity in the horizontal equation? (Besides,if it were, it would be negative).

In the yf=vyt+.5gt^2 , the last term should be negative (-.5gt^2). Upwards is postive, downward negative.

let V = initial velocity

xf =xi + vxi t

There is no acceleration in the horizontal direction, horizontal speed is constant!

vxi = V cos 37
so xf = 0 + .8 V t = 147
solve for V t

Now do y direction (up)
yf = yi + V t sin 37 - .5 (9.8) t^2
Note g is - if up is +
27 = 1 + V t sin 37 -4.9 t^2
we know V t
solve for t

and, on the yf, don't forget the initial height.

Yf=yi+Vy*t-1/2 g t^2

To solve this problem, we can break it down into components. Since we are given the height (27 m), distance (147 m), and angle (37°), we can calculate the initial velocity (v0) of the ball.

First, let's find the time it takes for the ball to reach the height of the wall. We can use the formula yf = yi + vit + 0.5gt^2, where yf is the final height, yi is the initial height, vi is the initial velocity, g is the acceleration due to gravity, and t is the time.

Here, yi = 1 m, yf = 27 m, g = 9.8 m/s^2. Plugging in these values, we get:

27 = 1 + v0 * sin(37°) * t - 0.5 * 9.8 * t^2

Simplifying further, we have:

26 = v0 * sin(37°) * t - 4.9 * t^2 ----(1)

Next, let's find the time it takes for the ball to travel the horizontal distance of 147 m. We can use the formula xf = xi + vixt, where xf is the final horizontal position, xi is the initial horizontal position, vix is the initial horizontal velocity, and t is the time.

Here, xi = 0 (since we are starting from home plate), xf = 147 m, and the ball is hit at an angle of 37°, so the initial horizontal velocity is v0 * cos(37°). Plugging in these values, we get:

147 = 0 + v0 * cos(37°) * t

Simplifying further, we have:

147 = v0 * cos(37°) * t ----(2)

Now, we have two equations (1 and 2) with two unknowns (v0 and t). We can solve this system of equations to find the values of v0 and t.

From equation (1), we can rearrange it as:

4.9 * t^2 - v0 * sin(37°) * t + 26 = 0

This is now a quadratic equation in terms of t. We can solve for t using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac))/(2a)

Here, a = 4.9, b = -v0 * sin(37°), and c = 26.

t = (-(-v0 * sin(37°)) ± sqrt((-v0 * sin(37°))^2 - 4 * 4.9 * 26))/(2 * 4.9)

Simplifying further, we have:

t = (v0 * sin(37°) ± √(v0^2 * sin^2(37°) - 4 * 4.9 * 26))/(9.8) ----(3)

Now, we can substitute this value of t from equation (3) into equation (2) to get an equation only in terms of v0:

147 = v0 * cos(37°) * ((v0 * sin(37°) ± √(v0^2 * sin^2(37°) - 4 * 4.9 * 26))/(9.8))

Simplifying further, we have:

147 = v0^2 * cos(37°) * sin(37°) ± √(v0^2 * sin^2(37°) - 4 * 4.9 * 26) * cos(37°) / (9.8)

Now, we can solve this equation to find the initial velocity (v0).