Online Homework

Projectile Motion Maximum Height

When a body is thrown with some initial velocity and is allowed to move only under the action of gravity is known as a projectile.The projectle moves in a parabolic path .This motion is called projectile motion.


Projectile motion maximum height- formula


The maximum vertical displacement attained by the projectile is known as the maximum height reached by the projectile.In the figure EA is the maximum height attained by the projectile motion.It is given by hmax.


  Projectile Motion


At point O, the initial velocity(u1) =usin`theta`

At point A, the final velocity (u3) = 0

The vertical distance travelled by the body is given by  (distance parallel to the y axis) sy = hmax

From the equations of motion, `u_3^2 = u_1^2-2gs_y`

Plugging in the known values , (0)2 = (usin`theta` )2 - 2ghmax

`rArr 2gh_(max) = u^2sin^2theta`

`rArr h_(max) = (u^2sin^2theta)/(2g)`

This is the equation for maximum height attained in a projectile motion.


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Time takien to attain maximun height - projectile motion


Let t be the time taken by the projectile to reach its maximum height.

From equation of motion u3 = u1 - gt

Plugging in the known values

0 = usin`theta` -gt

`rArr "gt" = u sintheta`

`rArr t = (usintheta)/g`


Projectile motion - maximum height , time taken - problems


Ex : 1 A body is projected upwards with velocity of 30 `m/sec` at  an angle of 300 with the horizontal . Determine the maximum height attained by the body and time taken to attain the maximum height.

Solution: Given velocity u = 30 `m/sec`

Angle with horizontal `theta` = 300

Acceleration due to gravity g="9.8 `m/sec^2`

Formulas used :

Maximum height attained is hmax `(u^2sin^2theta)/(2g)`

Time taken to attain maximum height t = `(usintheta)/g`

sin30 = `1/2`

Working : hmax = `(u^2sin^2theta)/(2g)`

`rArr`                  = `(30^2 (1/2)^2)/(2xx9.8)`

               hmax= `(900xx 1/4)/(19.6)` = 11.48 m

Time taken t =  `(usintheta)/g` = `(30xx1/2)/9.8` = 1.53 sec