Question

If a projectile is launched vertically upward from the ground with an initial velocity of 70 ft per​ sec, neglecting air​ resistance, its height s​ (in feet) above the ground t seconds after projection is given by s equals negative 16 t squared plus 70 t. Which equation should be used to determine the time at which the height of the projectile reaches 10 ​ft?

Answer 1

Answer:

Step-by-step explanation:

Given that a projectile is launched vertically upward from the ground with an initial velocity of 70 ft per​ sec

Neglecting air​ resistance, its height s​ (in feet) above the ground t seconds after projection is given by

$s=-16t^2+70t$

When height = 10 feet we have to find the time

Hence substitute s = 10 ft

$10=-16t^2+70t\\16t^2-70t+10=0$

We may get two answers as

$0.148, 4.227 seconds$

The first time is for going up and the second is for coming down.