A ball of mass 0.3 kg, initially at rest, is projected from ground level toward a wall that is 27.0 m away. The ball's velocity the moment it is projected is 75.0 m/s at 60° relative to the horizontal, and the wall is 11.0 m high. During its flight, the ball impacts nothing else and is not subjected to air resistance.
Vi = 75 sin 60 = 65 m/s
u = 75 cos 60 = 37.5 m/s
to go 27 m horizontal takes
t = 27/37.5 = .72 seconds
How high will it be (I assume that is the question) ?
v = Vi - 9.8 t
h = 0 + Vi t - 4.9 t^2
h at .72 s = 65(.72) - 4.9 (.72)^2
= 44.25 m high
Well, that's quite the projectile situation! Let's dive in and see what we can calculate.
First, let's break down the initial velocity into horizontal and vertical components. The horizontal component would be 75.0 m/s * cos(60°), giving us 75.0 * 0.5 = 37.5 m/s. The vertical component would be 75.0 m/s * sin(60°), giving us 75.0 * 0.866 = 64.95 m/s.
Now, let's calculate how long it takes for the ball to reach the wall. We'll use the horizontal component to do that. The formula for time is distance divided by velocity, so 27.0 m / 37.5 m/s = 0.72 seconds.
During this time, let's see how high the ball reaches. We can use the vertical component and the formula for projectile motion. Using the equation s = ut + 0.5at^2, where s is the displacement, u is the initial velocity, t is time, and a is acceleration, we can solve for the height.
In this case, the initial velocity is 64.95 m/s, the time is 0.72 seconds, and the acceleration is due to gravity (-9.8 m/s^2). Plugging in these values gives us s = (64.95 * 0.72) + (0.5 * -9.8 * (0.72)^2), which simplifies to s ≈ 27.8628 meters.
Finally, we can compare this height with the height of the wall (11.0 meters). Since the ball reached a height of 27.8628 meters, it definitely cleared the wall. Nice shot!
Just remember, in actual games, it's probably not a good idea to launch balls at walls unless you're playing some very unique version of baseball. Safety first!
To find the time it takes for the ball to reach the wall, we can use the horizontal component of the initial velocity.
Step 1: Find the horizontal velocity component (Vx).
The horizontal velocity (Vx) can be obtained by multiplying the initial velocity (V) by the cosine of the angle (θ).
Vx = V * cos(θ)
Vx = 75.0 m/s * cos(60°)
Step 2: Calculate the time for the ball to reach the wall.
The time (t) can be calculated by dividing the horizontal distance (d) by the horizontal velocity (Vx).
t = d / Vx
t = 27.0 m / (75.0 m/s * cos(60°))
Now, let's solve the equation:
Vx = 75.0 m/s * cos(60°)
Vx = 75.0 m/s * 0.5
Vx = 37.5 m/s
t = 27.0 m / (37.5 m/s)
t ≈ 0.72 s
Therefore, it takes approximately 0.72 seconds for the ball to reach the wall.
Now let's calculate the maximum height reached by the ball.
Step 1: Find the initial vertical velocity (Vy).
The initial vertical velocity (Vy) can be obtained by multiplying the initial velocity (V) by the sine of the angle (θ).
Vy = V * sin(θ)
Vy = 75.0 m/s * sin(60°)
Step 2: Calculate the time of flight (T).
The total time of flight (T) can be calculated using the equation:
T = 2 * t
T = 2 * 0.72 s
Step 3: Calculate the maximum height (h).
The maximum height (h) can be calculated using the equation:
h = Vy * T - (0.5 * g * T^2)
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, let's solve the equation:
Vy = 75.0 m/s * sin(60°)
Vy = 75.0 m/s * 0.866
Vy ≈ 64.95 m/s
T = 2 * 0.72 s
T = 1.44 s
h = 64.95 m/s * 1.44 s - (0.5 * 9.8 m/s^2 * (1.44 s)^2)
h ≈ 93.05 m
Therefore, the maximum height reached by the ball is approximately 93.05 meters.