A projectile is fired at 45.0° above the horizontal. Its initial speed is equal to 42.5 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?
Vi = 42.5 sin 45 = 30 m/s
v = Vi - 9.81 t
0 = 30 - 9.81 t
t = 3.06 seconds to top
h = 0 + Vi t - (1/2)9.81 t^2
h = 30(3.06) - 4.9 (3.06)^2
= 91.8 - 45.9
= 91.7 m
a ball thrown vertically upward from the top of a tower 60m high a velocity of 30ms what is the maximum height above the ground level.how long does it take to reach the ground
To find the maximum height reached by the projectile, we can use the kinematic equations of motion.
1. We can determine the vertical component of the initial velocity using the given data:
Initial velocity (v0) = 42.5 m/s
Angle above horizontal (θ) = 45.0°
Vertical component of initial velocity (v0y) = v0 * sin(θ)
v0y = 42.5 m/s * sin(45.0°)
v0y = 30.077 m/s (rounded to three decimal places)
2. Calculate the time taken to reach the maximum height:
Since the object reaches maximum height when its vertical velocity becomes zero, we can use the formula:
v = v0y - g * t
where:
v = final vertical velocity (zero at maximum height)
v0y = initial vertical velocity (30.077 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken to reach maximum height
0 = v0y - g * t
t = v0y / g
t = 30.077 m/s / 9.8 m/s^2
t ≈ 3.074 seconds (rounded to three decimal places)
3. Determine the maximum height:
The maximum height can be calculated using the equation:
h = v0y * t - (1/2) * g * t^2
where:
h = maximum height
v0y = initial vertical velocity (30.077 m/s)
t = time taken to reach maximum height (3.074 seconds)
g = acceleration due to gravity (9.8 m/s^2)
h = 30.077 m/s * 3.074 s - (1/2) * 9.8 m/s^2 * (3.074 s)^2
h ≈ 46.082 meters (rounded to three decimal places)
Therefore, the maximum height reached by the projectile is approximately 46.082 meters. It takes approximately 3.074 seconds to reach this height after being fired.
To find the maximum height reached by the projectile, we can analyze the motion in vertical direction separately from the horizontal motion.
First, let's find the time it takes for the projectile to reach its maximum height.
We know that the initial vertical velocity (Vy) is given by Vy = V * sin(θ), where V is the initial speed (42.5 m/s) and θ is the launch angle (45.0°).
Since the projectile reaches its peak when the vertical velocity becomes zero, we can use the equation of motion in the vertical direction to find the time it takes to reach the maximum height:
Vy = Vy0 + (a * t)
0 = Vy - (g * t) (where Vy0 is the initial vertical velocity and a is the vertical acceleration)
Substituting the values into the equation, we have:
0 = V * sin(θ) - (g * t)
Simplifying the equation, we get:
t = V * sin(θ) / g
Now, we can calculate the time it takes for the projectile to reach its maximum height using the given values:
t = (42.5 m/s) * sin(45.0°) / (9.8 m/s^2) ≈ 3.10 s
Next, we can find the maximum height reached by the projectile.
To calculate the maximum height (h), we can use the equation of motion in the vertical direction:
h = Vy0 * t + (1/2) * a * t^2
Since the projectile reaches its maximum height when its vertical velocity becomes zero, we can simplify the equation as follows:
h = (Vy0^2) / (2 * g)
Substituting the values:
h = (42.5 m/s * sin(45.0°))^2 / (2 * 9.8 m/s^2) ≈ 92.9 m
Therefore, the maximum height reached by the projectile is approximately 92.9 meters.
Note: This solution assumes that the launch height is the same as the landing height and that the effects of air resistance are negligible.