A projectile is fired at time t = 0.0 s, from point 0 at the edge of a cliff, with initial velocity components of V ax= 50m/s and V ay= 200m/s. The projectile rises, then falls into the sea at point P. The time of flight of the projectile is 50.0s What is the y-coordinate of the projectile when its x-coordinate is 1000 m?

As we know we can split projectile trajectory in two parts and the time for each part is equal. We can conclude that time t=20s belong to first part where projectile is going up.

Formula for this type of movement is y=Vay-(gt∧2)/2

If we take acceleration of gravity g=10m/s∧2

we get y=200*20-(10*20∧2)/2=4000-2000=2000m

But if I calculate time it takes the highest point I get

t=Vay/g=200/10=20s

The highest point for that time is

y= H=Vay∧2/2g=200∧2/2*10=40000/20=2000m

This happened because the projecile is fired from the edge of a cliff which is above the sea line.

Good luck!!!

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-07T11:37:50+02:00

The y-coordinate of the projectile is 2,040 m when x-coordinate is 1,000 m.

The given parameters;

initial horizontal component of the velocity, [tex]V_{ax}[/tex] = 50 m/s
initial vertical component of the velocity, [tex]V_{ay}[/tex] = 200 m/s
time of flight of the projectile, t = 50 s

The time of flight when x-coordinate is 1000 m is calculated as;

X = [tex]V_{ax}[/tex] t

1000 = 50t

t = 20 s

When the time of flight is 20 s, the y-coordinate is calculated as;

y = [tex]V_{ay}[/tex]t - ¹/₂gt²

y = 200 x 20 - (0.5 x 9.8 x 20²)

y = 4000 - 1960

y = 2,040 m

Thus, the y-coordinate of the projectile is 2,040 m when x-coordinate is 1,000 m.

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