Thread: Initial velocity of projectile launced from above origin

1. Initial velocity of projectile launced from above origin

I understand how to calculate the initial velocity of an object launched at an angle from the normal, but suppose i have a projectile launched at a 45 degree angle 10 meters above the X-axis and the projectile lands on the X-axis 100 meters from where it was launched. How do I calculate the initial velocity of the projectile in this hypothetical?

2. Originally Posted by Mr. Rogers
I understand how to calculate the initial velocity of an object launched at an angle from the normal, but suppose i have a projectile launched at a 45 degree angle 10 meters above the X-axis and the projectile lands on the X-axis 100 meters from where it was launched. How do I calculate the initial velocity of the projectile in this hypothetical?
Motion in the horizontal direction:

Initial velocity $= \frac{u}{\sqrt{2}}$ m/s.
Acceleration $= 0 ~ \text{m/s}^2$.
Displacement = 100 m
Therefore t = ..... (A)

Motion in the vertical direction (take upwards as positive and take the origin to be where the object is launched):

Initial velocity $= \frac{u}{\sqrt{2}}$ m/s.
Acceleration $= 9.8 ~ \text{m/s}^2$.
Displacement = -10 m.
t = expression found above (A).

Substitute all this into $x = ut + \frac{1}{2} a t^2$ and solve for u.