A projectile is launched from ground level at 37.0 m/s at an angle of 30.6 ° above horizontal. Use the launch point as the origin of your coordinate system.

(a) How much time elapses before the projectile is at a point 8.9 m above the ground and heading downwards toward the ground?
(b) How far downrange (the horizontal distance from the origin) was the projectile when it reached the highest point in its flight?

My answer for part a) 3.29 s
How do I get part b?

Vo = 37m/s[30.6o]

Xo = 37*cos30.6 = 31.8 m/s.
Yo = 37*sin30.6 = 18.8 m/s.

a. Yo*t + 0.5g*t^2 = 8.9 m.
18.8t - 4.9t^2 = 8.9
-4.9t^2 + 18.8t - 8.9 = 0
t = 3.28 s. and heading downward.
(Used Quad. formula).

b. Y = Yo - g*Tr = 0 at max ht.
18.8 - 9.8*Tr = 0
9.8Tr = 18.8
Tr = 1.92 s = Rise time or time to reach max ht.

Hor. Distance=Xo * Tr=31.8m/s * 1.92s =
61 m.

To find the horizontal distance covered by the projectile when it reaches the highest point in its flight, we can use the concept of symmetry in projectile motion.

Since the projectile is launched symmetrically with respect to the highest point, the time it takes to reach the highest point is half of the total flight time. Therefore, if we determine the total flight time, we can divide it by 2 to find the time it takes to reach the highest point.

To find the total flight time, we need to determine the time it takes for the projectile to reach the maximum height. The vertical motion of the projectile can be analyzed independently from its horizontal motion.

Given that the initial vertical velocity (Vy) is 37.0 m/s and the angle of projection (θ) is 30.6° above the horizontal, we can break down the initial velocity into vertical and horizontal components.

Vertical component of initial velocity (Vy) = 37.0 m/s * sin(30.6°)
Vertical component of initial velocity (Vy) = 18.88 m/s

Using this vertical component, we can determine the time it takes for the projectile to reach its maximum height using the kinematic equation:

Vy = Vy0 + at

Where Vy is the final vertical velocity, Vy0 is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s²), and t is the time.

At the highest point, the final vertical velocity Vy is zero, so we can rewrite the equation as:

0 = 18.88 m/s + (-9.8 m/s²) * t_max

Solving for t_max, we can find the time it takes for the projectile to reach its maximum height:

t_max = 18.88 m/s / 9.8 m/s²
t_max = 1.93 s

Now that we have the time it takes to reach the maximum height, we can calculate the total flight time:

Total flight time = 2 * t_max
Total flight time = 2 * 1.93 s
Total flight time = 3.86 s

Finally, to find the horizontal distance covered when the projectile reaches the highest point, we need to multiply the horizontal component of the initial velocity by the total flight time:

Horizontal distance = 37.0 m/s * cos(30.6°) * Total flight time

Plugging in the values:

Horizontal distance = 37.0 m/s * cos(30.6°) * 3.86 s

Calculating the above equation will give you the answer to part (b), the distance downrange when the projectile reaches its highest point.