A baseball player wants to hit a home run over the wall of a stadium He hits the ball 0888 m above the ground with at an angle of 30 degrees above the horizontal and a speed of 478 ms What is the tall

First, we need to find the maximum height the ball reaches. We can use the formula:

y = y0 + v0yt - 1/2g*t^2

where y is the vertical position, y0 is the initial vertical position (0.888 m), v0y is the initial vertical velocity (47.8*sin(30) m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time the ball is in the air.

To find the time the ball is in the air, we can use the formula:

t = 2v0sin(theta)/g

where theta is the angle above the horizontal (30 degrees). Plugging in the values, we get:

t = 247.8sin(30)/9.8 = 5.08 s

Now we can find the maximum height by plugging in the values:

y = 0.888 + 47.8sin(30)5.08 - 1/29.8(5.08)^2 = 34.1 m

Next, we need to find the horizontal distance the ball travels. We can use the formula:

x = v0x*t

where v0x is the initial horizontal velocity (47.8*cos(30) m/s). Plugging in the values, we get:

x = 47.8*cos(30)*5.08 = 203.4 m

Finally, we can use the Pythagorean theorem to find the distance from the wall to the point where the ball lands:

d = sqrt(x^2 + y^2) = sqrt(203.4^2 + 34.1^2) = 208.2 m

Since the wall is 90.9 m away, the player cannot clear it