Practice Problem 3.06 - Ennanced - with Feedback 11 of 11 Review | Constants In this example we will look at the more complex question of whether or not a projectile will clear a certain height at a specific point in its trajectory. Consider a field goal attempt, where a football is kicked from a point on the ground that is a horizontal distance d from the goalpost. For the attempt to be successful, the ball must clear the crossbar, 10 ft (about 3.05 m) above the ground, as shown in (Figure 1). The ball leaves the kicker's foot with an initial speed of 20.0 m/s at an angle of 30.0 above the horizontal. What is the distance d between kicker and goalpost if the ball barely clears the crossbar on its way back down? SOLUTION SET UP We use our idealized model of projectile motion, in which we assume level ground, ignore the effects of air resistance, and treat the football as a point particle. We place the origin of coordinates at the point where the ball is kicked. Then x0 = y0 = 0, Vox Voy = = vo cos 30.0 (20.0 m/s) (0.866) = 17.3 m/s, and (20.0 m/s) (0.500) = 10.0 m/s vo sin 30.0 = SOLVE We first ask when (i.e., at what value of t) the ball is at a height of 3.05 m above the ground; then we find the value of x at that time. When that value of x is equal to the distance d, the ball is just barely passing over the crossbar. To find the time t when y = 3.05 m, we use the equation y = y + voyt - gt. Substituting numerical values, we obtain 3.05 m = (10.0 m/s)t (4.90 m/s)t 1 2 2 . This is a quadratic equation; to solve it, we first write it in standard form: (4.90 m/s) t - (10.0 m/s)t + 3.05 m = 0 Then we use the quadratic formula. We get 1 t = 2(4.9 m/s) (10.0 m/s (10.0 m/s) - 4(4.90 m/s) (3.05 m) = 0.373 s, 1.67 s Figure 3.05 m Vo 20.0 m/s 30.0 1 of 1 There are two roots: t = 0.373 s and t = 1.67 s. The ball passes the height 3.05 m twice, once on the way up and once on the way down. We need to find the value of x at each of these times, using the equation for x as a function of time. Because x0 = 0, we have simply xvoxt. For t = 0.373 s, we get x = (17.3 m/s)(0.373 s) = = 6.45 m and for t = 1.67 s, we get , x = (17.3 m/s) (1.67 s) = 28.9 m . So, if the goalpost is located between 6.45 m and 28.9 m from the initial point, the ball will pass over the crossbar; otherwise, it will pass under it. REFLECT The distance of 28.9 m is about 32 yd; field goal attempts are often successful at that distance. The ball passes the height y = 10 ft twice, once on the way up (when t = 0.373 s) and once on the way down (when t = 1.67 s ). To verify this, we could calculate vy at both times; when we do, we find that it is positive (upward) at t = 0.373 s and negative (downward) with the same magnitude at t = 1.67 s. Part A - Practice Problem: The ball will have sufficient height to clear the goalpost at two different values of x. However, only the height at x = d describes the ball on its way back down. If the kicker gives the ball the same initial speed and angle but the ball is kicked from a point 24 m from the goalpost, what is the height of the ball above the crossbar as it crosses over the goalpost? Express your answer in meters. ? Pearson