A ball is thrown vertically upward with a speed of 1.86 m/s from a point 3.82 m above the ground.Calculate the time in which the ball will reach the ground.
To solve for t, use the formula:
`s=v_it + 1/2at^2`
where `s` - displacement `v_i` - initial velocity
`t` - time `a` - accelaration
Note that displacement refers to the change in height. So value of s is:
Also, acceleration is equal to gravity since the ball exhibits the principle of freely falling bodies. Its sign is negative because the direction of the intial velocity is upward. So, `a=-9.81 m/s^2` .
Then, substitute `s=-3.82` , `v_i=1.86` and `a=-9.82` .
`-3.82=1.86t + 1/2(-9.82)t^2`
Express the equation in quadratic form `ax^2+bx+c=0` .
Apply the quadratic formula to solve for t.
`t_1=1.09` and `t_2= -0.71`
Since t represents time, take only the positve value.
Hence, the time it takes for the ball to reach the ground is 1.09 seconds.