Question

A projectile 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

Answer 1

Answer:

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 $m/s^2$ 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 , $t=\frac{2usin\theta }{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$, 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

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

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

a) $H=\frac{u^2sin^2\theta}{2g}=\frac{192^2\times sin^290}{2\times 9.81}=1878.90m$

Maximum height reached = 1878.90 m

b) $t=\frac{2usin\theta }{g}=\frac{2\times 192\times sin90}{9.81}=39.14s$

Time of flight = 39.14 seconds.