A projectile is fired at 30.0° above the horizontal. Its initial speed is equal to 172.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile? At what time after being fired does the projectile reach this maximum height?

Question

A projectile is fired at 30.0° above the horizontal. Its initial speed is equal to 172.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile? At what time after being fired does the projectile reach this maximum height?

Answer 1

So in projectile problems horizontal a=0 so the velocity is constant. So finding x and y components for initial velocity

X: 172.5m/s*cos30=149.4m/s
Y: 172.5m/s*sin30=86.25m/s

The max height is at the apex of the projectile, so use vertical velocity in
Vfy =Viy+at---> -vi/a=t
-86.25m/s)/-9.8m/s^2=8.80 sec

Y=yo+vot+1/2at^2
Y= 87.25m/s*8.8s +.5*9.8m/s^2*8.8s^2
Y= 759m+379.456m
Y= 1138m high

Hope this helps!
[vfy=0 at apex]

Answer 2

Vo = 172.5m/s[30o].

Xo = 172.5*Cos30 = 149.4 m/s.
Yo = 172.5*sin30 = 86.25 m/s.

a. Yf^2 = Yo^2 + 2g*h.
h = -(Yo)^2/2g = -(86.25)^2/19.6 = 379.5 m.

b. Y = Yo + g*Tr = 0.
Tr = -Yo/g = -86.25/-9.8 = 8.80 s. = Rise time.

NOTE: In your Eq, use (-)9.8 for g.

Answer 3

To find the maximum height reached by the projectile, we need to first find the time it takes to reach the highest point of its trajectory. We can use the kinematic equation for vertical motion:

y = y0 + V0y * t - 1/2 * g * t^2

Where:
y = height (unknown)
y0 = initial height (0 in this case)
V0y = initial vertical component of velocity (V0 * sin(theta))
g = acceleration due to gravity (9.8 m/s^2)
t = time (unknown)

Since the projectile's initial vertical velocity is V0 * sin(theta), we can calculate it as:

V0y = 172.5 m/s * sin(30°) = 172.5 m/s * 0.5 = 86.25 m/s

Now, let's consider the motion at the highest point of the trajectory. At this point, the vertical velocity is momentarily zero. Therefore, we can write:

Vf = Viy - g * t

Where:
Vf = final velocity (0 m/s at the highest point)
Viy = initial vertical velocity (86.25 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time to reach the highest point (unknown)

Rearranging the equation, we get:

t = Viy / g

Substituting the values, we have:

t = 86.25 m/s / 9.8 m/s^2 ≈ 8.8 seconds

So, it takes approximately 8.8 seconds for the projectile to reach the highest point of its trajectory.

Now, let's find the maximum height reached by the projectile. We can use the same equation for vertical motion:

y = y0 + V0y * t - 1/2 * g * t^2

Substituting the values:

y = 0 + 86.25 m/s * 8.8 s - 1/2 * 9.8 m/s^2 * (8.8 s)^2

Simplifying,

y = 0 + 759 m - 1/2 * 9.8 m/s^2 * 77.44 s^2

y ≈ 0 + 759 m - 379.936 m

Therefore, the maximum height reached by the projectile is approximately:

y ≈ 379.064 meters

So, the maximum height reached is approximately 379.064 meters.

To find the time after being fired when the projectile reaches this maximum height, we can simply use the time it took to reach the highest point, which is approximately 8.8 seconds.