Degrees with the horizontal. The initial speed of the projectile is 50 m/s. Ignoring air resistance, find:
a) The time it takes for the projectile to reach its maximum height.
b) The maximum height reached by the projectile.
c) The total time of flight of the projectile.
d) The horizontal range of the projectile.

To solve this problem, we can break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes due to the effect of gravity.

a) To find the time it takes for the projectile to reach its maximum height, we need to find the time it takes for the vertical component of the velocity to become zero. We can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the final velocity is zero when the projectile reaches its maximum height, so v = 0. The initial velocity in the vertical direction is given by:

u_y = u * sin(θ)

where θ is the angle with the horizontal. In this case, θ = 60 degrees.

Using the equation v = u + at, we have:

0 = u_y + (-g)t

where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.

Substituting the values, we have:

0 = (50 m/s) * sin(60 degrees) - (9.8 m/s^2) * t

Simplifying the equation, we get:

t = (50 m/s) * sin(60 degrees) / (9.8 m/s^2)

t ≈ 2.55 s

Therefore, it takes approximately 2.55 seconds for the projectile to reach its maximum height.

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

s = u_y * t + (1/2) * (-g) * t^2

where s is the displacement, t is the time, u_y is the initial vertical velocity, and g is the acceleration due to gravity.

Substituting the values, we have:

s = (50 m/s) * sin(60 degrees) * (2.55 s) + (1/2) * (-9.8 m/s^2) * (2.55 s)^2

Simplifying the equation, we get:

s ≈ 65.1 m

Therefore, the maximum height reached by the projectile is approximately 65.1 meters.

c) The total time of flight of the projectile can be found by doubling the time it takes for the projectile to reach its maximum height:

Total time of flight = 2 * (2.55 s) = 5.1 s

Therefore, the total time of flight of the projectile is 5.1 seconds.

d) The horizontal range of the projectile can be found using the equation:

Range = u_x * total time of flight

where u_x is the initial horizontal velocity.

The initial horizontal velocity can be found using:

u_x = u * cos(θ)

Substituting the values, we have:

Range = (50 m/s) * cos(60 degrees) * (5.1 s)

Simplifying the equation, we get:

Range ≈