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swat32
3 years ago
8

A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same inst

ant another ball B is thrown upward from the ground with an initial velocity of 20 m>s. Determine the height from the ground and the time at which they pass.
Engineering
1 answer:
expeople1 [14]3 years ago
6 0

Answer:

s= 20.4 m  

Explanation:

First lets write down equations for each ball:  

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:  

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b  

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s  

now we just have to put that time in any of those equations an get distance from the ground:  

s = 30 + 5*2 -1/2*9.81 *2^2  

s= 20.4 m  

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The critical length of surface flaw = 6.176 mm

Explanation:

Given data-

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Answer:

#include <iostream>

#include <iomanip>

#include <string>

using namespace std;

int main() {

   string name[5];  

   int age[5];  

   int i,j;  

   for ( i = 0; i<=4; i++ ) {  

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