A projectile is fired at 12.5m/s at angle of 53.1,with the horizontal from apoint 75.0 m above the ground.(a) how lonv it take to reach the ground?(b)what maximum height does it reach?

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

(a) To find the time it takes for the projectile to reach the ground, we can use the equation:

y = y0 + v0y * t - (1/2) * g * t^2

where:
y = final vertical position (0 m, since it reaches the ground)
y0 = initial vertical position (75.0 m)
v0y = initial vertical velocity (v0 * sin(theta))
g = acceleration due to gravity (9.8 m/s^2)
t = time

Plugging in the values, we have:

0 = 75.0 + (12.5 * sin(53.1)) * t - (1/2) * 9.8 * t^2

Rearranging the equation, we get:

4.9 * t^2 - (12.5 * sin(53.1)) * t - 75.0 = 0

Solving this quadratic equation, we find two possible values for t: t = 0.0 s and t = 5.08 s.

Since time cannot be negative, the projectile takes approximately 5.08 seconds to reach the ground.

(b) To find the maximum height reached by the projectile, we can use the equation:

vfy = v0y - g * t

where:
vfy = final vertical velocity (0 m/s, at the maximum height)
v0y = initial vertical velocity (v0 * sin(theta))
g = acceleration due to gravity (9.8 m/s^2)
t = time to reach the maximum height (half of the total time of flight)

Plugging in the values, we have:

0 = (12.5 * sin(53.1)) - 9.8 * (5.08 / 2)

Simplifying the equation, we find:

0 = 6.62 - 24.98

This equation is not satisfied, which means that the projectile does not reach a maximum height. Instead, it reaches the ground directly.