Question

An object is hit w/ a resultant velocity of 50 m/s at an angle of 30 degrees above horizontal.

Calculate the time from the highest point to the ground.

If the ball is hit and raised 10 meters above the grass, how far will it travel?

Answer 1

To tackle the first problem, let's first figure out how high it goes, and then we can figure out how long it takes to fall from that highest point. If it has an initial velocity \(v_i\) and initial angle \(\theta\), then the velocity in the vertical direction is given by \(v_i \sin \theta\). We will now use the kinematic equation \(v_f^2 = v_i^2 + 2\vec a \cdot \Delta \vec r\) to figure out how far it goes. Take note that the acceleration is downward so \(a=-g\) and the final velocity is \(v_f=0\) since we are considering the interval where it comes to a stop with respect to the vertical component of velocity at the end (the peak). We will rename \(\Delta \vec r\) as \(\Delta y\) to remind us we are talking about the vertical direction only. \[\underbrace{0^2=(v_i\sin\theta)^2-2g\Delta y }_{v_f^2 = v_i^2 + 2\vec a \cdot \Delta \vec r}\Rightarrow \Delta y = \frac{v_i^2 \sin^2 \theta}{2g}\]Now, let us use this height to find out how long it takes for the object to come back down. We will use the equation \(\Delta \vec r = \vec v_i t + \frac{1}{2}\vec at^2\). Similarly, \(\Delta \vec r\) becomes \(-\Delta y\) (negative because we're going downward), \(\vec a\) becomes \(-g\), and \(v_i=0\) because we are considering the interval where the vertical component of velocity is zero initially (this is DIFFERENT than the \(v_i\) from above since this is a new time interval).\[\underbrace{\underbrace{-\frac{v_i^2 \sin^2 \theta}{2g}}_{-\Delta y}= 0t - \frac{1}{2}gt^2}_{\Delta \vec r = \vec v_it + \frac{1}{2}\vec at^2}\Rightarrow \boxed{\displaystyle t=\frac{v_i\sin\theta}{g}}\]

Answer 2

For the second problem, consider the following. Again, the initial vertical velocity will be given by just \(v_i\sin\theta\). Let us consider again the equation \(\Delta \vec r = \vec v_it + \frac{1}{2}\vec a t^2\). Since the ball is raised by a distance \(\Delta y\) (given), we know that through the trajectory it must FALL by that amount, so thus \(\Delta \vec r = -\Delta y\). Additionally, the gravitational acceleration is \(\vec a =-g\). To solve for \(t\), the time it takes to travel through the parabola, we must use the quadratic formula.\[\underbrace{-\Delta y = v_it \sin \theta - \frac{1}{2}gt^2}_{\Delta \vec r = \vec v_it + \frac{1}{2}\vec a t^2} \Rightarrow \frac{g}{2}t^2-(v_i\sin\theta)t-\Delta y=0\]\[t=\frac{v_i\sin\theta \pm \sqrt{v_i^2 \sin^2 \theta + 2g\Delta y}}{g}\]We will omit the negative solution (when \(\pm\) is \(-\)) since we are only considering the future.\[t=\frac{v_i\sin\theta + \sqrt{v_i^2 \sin^2 \theta + 2g\Delta y}}{g}\]Now, since the horizontal velocity, \(v_i cos \theta\), is constant, we can find the horizontal displacement through this length of time \(\Delta x\) by evaluating \(\Delta x = (v_i \cos \theta) \cdot t\).\[\boxed{\displaystyle \Delta x = v_i \cos \theta \left(\frac{v_i\sin\theta + \sqrt{v_i^2 \sin^2 \theta + 2g\Delta y}}{g}\right)}\]It doesn't look pretty, but it is what it is.