Question

A ball is thrown directly downward with an initial speed of 7.15 m/s, from a height of 30.2 m. After what time interval does it strike the ground?

Answer 1

We are given the following information.

Downward motion.

Initial speed = -7.15 m/s

Change in height = -30.2 m

Acceleration due to gravity = -9.81 m/s^2

We are asked to find the time it takes the ball to strike the ground.

Recall from the equations of motion,

`\Delta y=v_0t+(1)/(2)gt^2`

Let us substitute the known values and solve for t.

`\begin{gathered} -30.2=-7.15t+(1)/(2)(-9.81)t^2 \\ -30.2=-7.15t-4.905t^2 \\ 4.905t^2+7.15t-30.2=0 \end{gathered}`

As you can see, it is a quadratic equation. We can use the quadratic formula to find the values of t.

`t=(-b\pm√(b^2-4ac))/(2a)`

The coefficients are

a = 4.905

b = 7.15

c = -30.2

`\begin{gathered} t=(-7.15\pm√(7.15^2-4(4.905)(-30.2)))/(2(4.905)) \\ t=(-7.15\pm25.37)/(9.81) \\ t=(-7.15+25.37)/(9.81),\;(-7.15-25.37)/(9.81) \\ t=1.86,\;-3.32 \end{gathered}`

Time cannot be negative, discard the negative value and accept the positive value.

Therefore, the time it takes for the ball to strike the ground is 1.86 seconds.