maximum height from the tower, we can use the equations for the motion of a projectile, which describe the horizontal and vertical components of the motion separately.
The horizontal component of the motion is given by the equation:
x = vt * cos(θ)
where x is the horizontal distance traveled, v is the speed, t is the time, and θ is the angle of projection.
The vertical component of the motion is given by the equation:
y = vt * sin(θ) - (1/2)gt^2
where y is the vertical distance traveled, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
In this case, the initial height of the projectile is 50 meters, so we can set y = 50 at t = 0. This gives us the following equation:
50 = (5 * 0) * sin(20) - (1/2) * 9.8 * 0^2
50 = 0
This equation is true for all values of t, so it does not provide any useful information.
To find the maximum height, we can set the derivative of y with respect to t equal to 0 and solve for t:
dy/dt = 0 = 5 * sin(20) - 9.8t
9.8t = 5 * sin(20)
t = 5 * sin(20) / 9.8
Substituting this value for t into the equation for y, we get:
y = (5 * sin(20)) / 9.8 - (1/2) * 9.8 * ((5 * sin(20)) / 9.8)^2
y = (5 * sin(20)) / 9.8 - (1/2) * 9.8 * (25 * sin^2(20)) / (9.8)^2
y = (5 * sin(20)) / 9.8 - (25 * sin^2(20)) / 9.8
y = (5 - 25 * sin^2(20)) / 9.8
To find the maximum value of y, we can take the derivative of this expression with respect to 20 and set it equal to 0:
dy/dθ = 0 = -50 * sin(20) * cos(20) / (9.8)
-50 * sin(20) * cos(20) = 0
sin(20) = 0 or cos(20) = 0
Since 20 is not a multiple of 90 degrees, we can disregard the solution sin(20) = 0. Therefore, cos(20) = 0, which means that θ = 90 degrees.
Substituting θ = 90 degrees into the expression for y, we get:
y = (5 - 25 * sin^2(90)) / 9.8
y = (5 - 25 * 1) / 9.8
y = (5 - 25) / 9.8
y = -20 / 9.8
y = -2.04
Since the initial height of the projectile is 50 meters, the maximum height above the tower is 50 - (-2.04) = 52.04 meters.