Consider the motion of a projectile. It is fired at t = 0. Its initial speed is 17 m/s and its initial projection angle is 35◦ from the horizontal.
What is the maximum height of its tra- jectory? The acceleration due to gravity is 9.8 m/s2 .
Answer in units of m.
vertical problem only
Vi = 17 sin 35 = 9.75
initial kinetic energy = (1/2) mVi^2
potential energy at top = m g h
so
g h = 1/2 Vi^2
h = VI^2/(2 g) = 9.75^2/(2*9.8) = 4.85
pathetic projectile :(
the initial vertical velocity is ... 17 m/s * sin(35º)
h = -4.9 t^2 + [17 * sin(35º)] t
max h is on the axis of symmetry ... t = [-17 sin(35º)] / (2 * -4.9)
thank you guys!
Well, let's have some fun with projectile motion, shall we? So, we have a projectile being launched with an initial speed of 17 m/s and an angle of 35 degrees. We want to find the maximum height of its trajectory.
Now, to find the maximum height, we need to break the projectile motion into two parts: the horizontal and vertical components. Since the projectile is not influenced by any forces horizontally (besides air resistance, but let's ignore that for now because we're having a good time), the horizontal component of velocity remains constant throughout the motion.
So, first, let's find the time it takes for the projectile to reach its maximum height. We can use the vertical component of velocity for that. The initial vertical velocity is calculated by multiplying the initial speed by the sine of the launching angle. In this case, it will be 17 m/s * sin(35 degrees).
Next, we use the formula for time of flight, which is given by t = (2 * initial vertical velocity) / g, where g is the acceleration due to gravity. In this case, it will be (2 * (17 m/s * sin(35 degrees))) / 9.8 m/s^2. Crunch the numbers, and you'll find the time it takes to reach the maximum height.
Now that we have the time, we can find the maximum height using the equation h = (initial vertical velocity)^2 / (2 * g). Plug in the values, and you'll get your answer. And make sure to give the answer in meters, because clowns never skip a beat in the metric system!
Have a blast with your projectile motion calculations! Hope this answer didn't go over your head!
To find the maximum height of the projectile's trajectory, we can use the equations of motion for projectile motion. The key is to break down the motion into horizontal and vertical components.
The initial speed of the projectile can be split into its horizontal and vertical components using trigonometry. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.
The horizontal component of the initial velocity (Vx) can be found using the equation:
Vx = V * cos(θ)
where V is the initial speed and θ is the projection angle.
In this case, V = 17 m/s and θ = 35 degrees, so:
Vx = 17 * cos(35)
Next, we need to find the time it takes for the projectile to reach its maximum height. At the highest point of its trajectory, the vertical component of its velocity will be zero. We can use the equation:
Vy = V * sin(θ) - g * t
where Vy is the vertical component of the velocity, g is the acceleration due to gravity, and t is the time.
At the maximum height, Vy will be zero, so we can rewrite the equation as:
0 = V * sin(θ) - g * t
We can rearrange this equation to solve for t:
t = V * sin(θ) / g
Substituting the given values:
t = 17 * sin(35) / 9.8
Now, we can find the maximum height (h) of the projectile using the equation:
h = Vy * t - (1/2) * g * t^2
Since Vy at the maximum height is 0, the equation simplifies to:
h = -(1/2) * g * t^2
Substituting the values we calculated:
h = -(1/2) * 9.8 * (17 * sin(35) / 9.8)^2
Evaluating this expression will give the maximum height of the projectile's trajectory in meters.