A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 177 feet per second from an initial height of 133 feet off the ground, then the height of the projectile, h, in feet, t seconds after it's shot is given by the equation:

h=−16t^2+177t+133

Find the two points in time when the object is 172 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).

Need on this geometry question. explain how you did it.

Solve for n? [tex]z = \frac{3}{4} mn[/tex]

Aspacecraft can attain a stable orbit 300 kilometers above earth if it reaches a velocity of 7.7 kilometers per second. the formula for a rocket's maximum velocity v in kilometers per second is vequalsminus0.0098tplusc ln upper r, where t is the firing time in seconds, c is the velocity of the exhaust in kilometers per second, and r is the ratio of the mass of the rocket filled with fuel to the mass of the rocket without fuel. find the velocity of a spacecraft whose booster rocket has a mass ratio of 20, an exhaust velocity of 2.1 km/s, and a firing time of 15 s. can the spacecraft achieve a stable orbit 300 km above earth?