Problema Solution
A ball is thrown upwards from the ground level withan initial velocity of 108 ft per. second. It's height h in feet after the seconds is given by the equation h= - 16t2 = =+ 108t. At what time will the ball hit the ground ?
Answer provided by our tutors
The initial velocity of the ball is 108 ft per second
Using;
V=u+gt
When the ball goes up the final velocity(v) will be zero and the acceleration due to gravity (g) whose value is 32.174 ft per second will act in opposite direction.
Hence the equation becomes
0=108-32.174*t
T(1)=3.357 seconds
In this time the ball reaches its maximum height.
Now h=108*t and intial velocity of ball is 0 during its return flight and using formula-
H=u*t+(g*t^2)/2
Substituting the values I get
108*t=(32.174*t^2)/2
Solving I get
T(2)=6.713
So total time in which the ball hits the ground is= t(1)+t(2)=3.357+6.713=10.070 seconds