Question

Example 4.3 A Bull's-Eye Every Time In a popular lecture demonstration, a projectile is fired at a target in such a way that the projectile leaves the gun at the same time the target is dropped from rest. Show that if the gun is initially aimed at the stationary target, the projectile hits the falling target as shown in figure (a). The velocity of the projectile (red arrows) changes in direction and magnitude, but its acceleration (purple arrows) remains constant. SOLVE IT Conceptualize We conceptualize the problem by studying figure (a). Notice that the problem does not ask for numerical values. The expected result must involve an algebraic argument. a Line of sight Categorize Because both objects are subject only to gravity, we Target categorize this problem as one involving two objects in free fall, ge the target moving in one dimension and the projectile x tan 8 moving in two. The target Tis Point of modeled as a particle under YT Gun 0 collision constant acceleration in one X1 dimension. The projectile P is adalad modeled as a particle under constant acceleration in the die y direction and a particle under b constant velocity in the (a) Multiflash photograph of the projectile-target demonstration. If the gun is x direction. aimed directly at the target and is fired at the same instant the target begins to fall, the projectile will hit the target. (b) Schematic diagram of the projectile- target demonstration. Analyze Figure (b) shows that the initial y coordinate y of the target is x tan 6, and its initial velocity is zero. It falls with acceleration a, = -9. (1) Write an expression for the y coordinate of the target at any moment after release, noting that its initial velocity is zero: Y = Yr + (0) - 2+ = xtan 8,- Write an expression for the y coordinate of the projectile at any moment: (2) Yo = Yip + Vypt - 19 = 0 + (vip sin 8.) - $t = (vip sin 8,)t - Write an expression for the x coordinate of the projectile at any moment: Xp = xp + vypt = 0 + (vp cost = (vecos et Solve this expression for time as a function of the horizontal position of the projectile: p Cose Substitute this expression into Equation (2): (3) vp = (vp sin 8.1(v.265 ) - 25+ = x, tan 8, - Finalize Compare Equations (1) and (3). We see that when the x coordinates of the projectile and target are the same-that is, when 1 = xp, their y coordinates given by Equations (1) and (3) are the same and a collision results. MASTER IT HINTS I'M STUCK! A projectile is fired at t = 0 with an initial velocity of v, at an angle of with respect to the horizontal towards the vertical side of a building that is a distance d away. (a) Determine an equation for the time at which the projectile strikes the building. (Use the following as necessary: d, v, and ) — X (b) Determine an equation for the height h above the ground at which the projectile strikes the building. (Use the following as necessary: d, v, 9, and 6) h = x

Answer 1

We have no acceleration due to gravity in horizontal translation of the projectile hence, time taken by projectile to cover distance d is simply

$t=\frac{d}{v_{i_x}}\Rightarrow \frac{d}{v_i\cos \theta}$

Using normal kinematic equations and applying the above obtained expression for time we have

$\\ h=v_i\sin \theta . t-\frac{1}{2}gt^2 \\ h=v_i \sin \theta. \left (\frac{d}{v_i\cos \theta} \right )-\frac{1}{2}g.\left (\frac{d}{v_i\cos \theta} \right )^2 \\ h=d\tan \theta -\frac{1}{2}g. \left (\frac{d}{v_i\cos \theta} \right )^2$