Aprojectile is fired from a gun near the surface of earth. the initial velocity of the projectile has a vertical component of 90 meters per second and a horizontal component of 49 m/s. how long will it take the projectile to reach the highest point in its path?
a100.s b20.s c10.s
d5.s

For this we only need to observe vertical component. Vertical component is 90 meters/s
The gravity is decresing vertical component at rate of 9.80 m/s because gravity acceleration is 9,80 m/s^2

Simply by deviding starting vertical speed and gravitational acceleration we get time required for  projectile to stop gaining altitude.

90/9.8 = 9,18s

Looking at the answers there isn't 9.18 offered as answer. But it is indeed answer for the question that has been asked. Maybe you thought that 49m/s is vertical speed than the answer would be 49/9.8=5

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward