Question

A projectile is launched vertically with an initial velocity of 50 m/s. How long would it take to reach the top of the launch?

Answer 1

It'd take it 5.10 seconds

Step-by-step explanation:

At the top of the launch, it is maximum height, h

formular derivation :

At maximum height, final velocity, v is zero

From first newtons equation:

`{ \sf{v = u + gt}}`

v is the final velocity

u is the initial velocity

g is acceleration due to gravity

t is the time taken

but v is zero:

`0 = u + gt \\ gt = u \\ t = (u)/(g)`

since it is vertical motion, u = usin(theta), where angle is 90°

`{ \boxed{t = (u \sin( \theta) )/(g) }}`

t is the time taken to reach the top of the launch:

substitute, taking g to be 9.8 m/s²

`t = (50 * \sin(90 \degree) )/(9.8) = (50)/(9.8) \\ \\ { \boxed{ \boxed{time = 5.10 \: seconds}}}`

Answer 2

Step-by-step explanation:

Givens

vi = 50 m/s

vf = 0

a = - 9.81

t = ?

Formula

t = (vf - vi)/a

Solution

t = (0 - 50)/-9.81

t = 5.097

Notice that a is made into a minus. It does not mean subtract. It means that the direction of the projectile and gravity must go in opposite directions. You will find that direction is a critically important consideration in physics.