A projectile is fired at t = 0 with an initial velocity of vi at an angle of ? w

Question

A projectile is fired at t = 0 with an initial velocity of vi at an angle of ? with respect to the horizontal towards the vertical side of a building that is a distance d away as shown in the figure. (a) Determine an equation for the time at which the projectile strikes the building. (Use any variable or symbol stated above as necessary.) t = s (b) Determine an equation for the height h above the ground at which the projectile strikes the building. (Use any variable or symbol stated above along with the following as necessary: g.) h = m

Explanation / Answer

Horizontal velocity remains constant throughout and = vi cos

Initial vertical velocity = vi sin

Let the horizontal distance of the building from point of fire be R.

Now time of flight, t = R / vi cos, because it depends only on horizontal velocity (considering that it hits the building).

Now considering downward motion as positive direction, initial vertical velocity = -vi sin.

Acceleration = +g

Using, s = ut + (1/2)at2,

height at which it hits the building, h = |-vi sin t + (1/2) g t2| = |-Rtan + (g R2) / (2vi2cos2) |, here i have put modulus because we have considered upward as negative direction and height above ground has to be positive (considering it hits the building).

And finally, to check whether it hits the building we would need to calculate the time after which the projectile would have come back to ground had there been no building. For this, only vertical velocity comes into play.

tv = (vi sin) / g,

is the time it takes to reach the highest point (using v = u + at).

So total time of flight, tf = 2(vi sin) / g

Now for the projectile to hit the building,

tf > t (t = R / vi cos)

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