srodr1110
contestada

A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16t^2 +v0t. Find the time(s) that the projectile with (a) reach a height of 64 ft and (b) return to the ground when v0 = 80 feet per second

Respuesta :

merlynthewhizz
The formula is [tex]s=-16t^2+V_ot[/tex]

We have:
[tex]s=64[/tex]
[tex]V_o=80ft/s[/tex]

Substituting these values into the formula, we have

[tex]64=-16t^2+80t[/tex]

[tex]16t^2-80t+64=0[/tex] ⇒ divide each term by 16

[tex]t^2-5t+4=0[/tex] ⇒ find the pair of numbers that multiply to give 4 and sum gives -5 which is -1 and -4

[tex](t-1)(t-4)=0[/tex] ⇒ We have two values of t

[tex]t-1=0 [/tex] ⇒ [tex]t=1[/tex]
[tex]t-4=0[/tex] ⇒ [tex]t=4[/tex]

The time when the projectile reached 64 feet is at t=1 and t=4



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