Respuesta :
t = 4 sec
Solution:
Given data:
Initial height ([tex]h_0[/tex]) = 24 feet
Initial velocity ([tex]v_0[/tex]) = 58 feet
To find the time taken by the object to hit the ground.
General formula for the height(feet) of a projectile over time:
[tex]h(t)=-16 t^{2}+v_{0} t+h_{0}[/tex]
Substitute the given data in the above formula.
[tex]h(t)=-16 t^{2}+58 t+24[/tex]
This equation looks like a quadratic equation.
We can solve it by using the quadratic formula:
[tex]$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Here a = –16, b = 58, c = 24
[tex]$t=\frac{-58 \pm \sqrt{(58)^{2}-4(-16)(24)}}{2(-16)}[/tex]
[tex]$=\frac{-58 \pm \sqrt{3364+1536}}{-32}[/tex]
[tex]$=\frac{-58 \pm \sqrt{4900}}{-32}[/tex]
[tex]$=\frac{-58 \pm 70}{-32}[/tex]
[tex]$t=\frac{-58 - 70}{-32} \ \text{(or)} \ t=\frac{-58 + 70}{-32}[/tex]
[tex]$t=\frac{-128}{-32} \ \text{(or)} \ t=\frac{12}{-32}[/tex]
[tex]$t=4 \ \text{(or)} \ t=-0.375[/tex]
Time cannot be indicated in negative. So neglect t = –0.375.
Therefore, t = 4 sec
Hence 4 sec taken by the object to hit the ground.
