Question

A projectile is launched vertically upward with an initial velocity of 128 ft/s.A) when will the projectile return to the ground?B) how high will it go?C) when will the projectile be 240 feet above ground?Use the formula: h= vt -16t^2

Answer 1

Given,

The initial velocity of the projectile, u=128 ft/s

The acceleration due to gravity, g=-32 ft/s²

After reaching the maximum height the velocity of the projectile will be v=0 m/s

The time(t₁) required for the projectile to reach the maximum height is given by,

`v=u+gt_1`

On substituting the known values,

`\begin{gathered} 0=128+(-32)* t_1 \\ t_1=(-128)/(-32) \\ =4\text{ s} \end{gathered}`

The maximum height (h) reached by the projectile is given by,

`v^2-u^2=2gh`

On substituting the known values,

`\begin{gathered} 0-128^2=2*-32* h \\ h=(-128^2)/(2*-32) \\ =256\text{ ft} \end{gathered}`

The time (t₂) required for the projectile to come back to the ground from its maximum height is calculated using the formula,

`h=vt_2-16t^2_2`

Here the value of the height will take a negative value because the projectile will be covering that distance when it is falling downwards

On substituting the known values,

`\begin{gathered} -256=0-16t^2_2 \\ \Rightarrow t_2=\sqrt{(256)/(16)} \\ =4\text{ s} \end{gathered}`

Thus the total time in which the projectile returns to the ground is

`\begin{gathered} t=t_1+t_2 \\ =4+4=8\text{ s} \end{gathered}`

A) Thus the projectile will return to the ground in 8 s

B) The maximum height reached by the projectile is 256 ft

C)

The height, h=240 ft

The time(t₃) required to reach this height is given by,

`h=ut_3-16t^2_3`

On substituting the known values,

`\begin{gathered} 240=128* t_3-16t^2_3 \\ \Rightarrow16t^2_3-128t_3+240=0 \end{gathered}`

On solving the above equation,

`t_3=5_{}\text{ s}`

And

`t_3=3\text{ s}`

Thus the projectile will reach the height at 3s when going up and again at 5s when coming down.