Question

In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m above the ground and drops a ball straight down. At the same moment, Amelia uses a spring-loaded device on the ground to launch a dart straight up toward the ball. The dart is launched at 16.2 m/s. Find the time and height of the collision by simultaneously solving the equations for the ball and the dart. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

Answer 1

Answer:

time of collision is

t = 0.395 s

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

$v_1 = 0$

similarly initial speed of the object which is projected by spring is given as

$v_2 = 16.2 m/s$

now relative velocity of object with respect to ball

$v_r = 16.2 m/s$

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

$d = v_r t$

$6.4 = 16.2 t$

$t = 0.395 m$

Now the height attained by the object in the same time is given as

$h = v_2 t - \frac{1}{2}gt^2$

$h = 16.2(0.395) - \frac{1}{2}(9.81).395^2$

$h = 5.63 m$

so they will collide at height of 5.63 m from ground