# Find the time of flight of a projectile thrown horizontally with speed 50 m/s from the top of an inclined plane which makes an angle of 45° with the horizontal.

*print*Print*list*Cite

### 1 Answer

The projectile is thrown in a horizontal direction at 50 m/s. It is thrown from the top of an inclined plane that slopes downwards at 45 degrees to the horizontal. This is described by the following diagram:

The red line is the inclined slope; and the projectile is released at 50 m/s from the top (open dot) in the direction of the arrow.

As the projectile moves forward it also moves downwards due to acceleration due to the gravitational attraction of the Earth. Let g = 10 m/s.

The projectile hits the slope when the distance traveled in the horizontal direction is equal to the distance traveled vertically downwards. Let the time taken by the projectile to hit the inclined plane be t. The distance traveled in the horizontal direction is 50*t. The distance traveled vertically down is (1/2)*10*t^2. As the slope is inclined downwards at 45 degrees,

50*t = (1/2)*10*t^2

=> 50 = 5*t

=> t = 10

The projectile's time of flight is 10 seconds.