1 2 P088 My In Example 2.12, two circus performers rehearse a trick in which a b
ID: 1879513 • Letter: 1
Question
1 2 P088 My In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 7.2 m above the ground and drops a ball straight down. At t spring-loaded device on the ground to launch a dart straight up toward the ball. The dart is launched at 19.6 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.) time height Need Help?Explanation / Answer
Solution:
Answer: The ball an dart meet after a time t = 0.367347 seconds
at a height of 6.53878 m above the ground
Let the ball and dart meet at some height h after a time t .
Ball has free fall ; so its initial velocity =v1i= 0 .
Dart has initial velocity ; so v2i = 19.6 m/s
Distance traveled by the ball to meet the dart is h = 1/2 gt^2 -------->(1)
Height thru which the dart went to meet the ball = 7.2 - h = vit+1/2gt^2
=> 7.2- h = 19.6 t -1/2gt^2 --------> (2)
substitute for h from equ(1) .
7.2 - 1/2gt^2 =19.6 -1/2gt^2
=>19.6 t = 7.2 => t = 7.2/19.6 =0.367 s
h =1/2gt^2 = 1/2 * 9.8 * (0.36735)^2 = 0.661 m
Height above the ground where they meet = 7.2 - h
= 7.2 - 0.66122 =6.53878 m
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