Aprojectile is fired vertically with an initial velocity of 192 m/s. calculate the maximum altitude h reached by the projectile and the time t after firing for it to return to the ground. neglect air resistance and take the gravitational acceleration to be constant at 9.81 m/s2.
answers:
h = m
t = s

a) Maximum height reached = 1878.90 m

b) Time of flight = 39.14 seconds.

Explanation:

Projectile motion has two types of motion Horizontal and Vertical motion.  

Vertical motion:  

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.  

Considering upward vertical motion of projectile.  

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.  

0 = u sin θ - gt  

t = u sin θ/g  

Total time for vertical motion is two times time taken for upward vertical motion of projectile.  

So total travel time of projectile ,

Vertical motion (Maximum height reached, H) :

We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.  

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

In the give problem we have u = 192 m/s, θ = 90° we need to find H and t.

a)

Maximum height reached = 1878.90 m

b)

Time of flight = 39.14 seconds.

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the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.