One projectile is launched from the side of a building which is 17 meters above

Question

One projectile is launched from the side of a building which is 17 meters above the ground. It is launched at 14 m/s at an angle of 29 degrees above the horizontal and lands on the ground after it travels some horizontal displacement (this would be how far it lands horizontally from the base of the building.) Another projectile is launched from the same height at the same speed, but at an angle of 29 degrees below the horizontal and lands on the ground after traveling some horizontal displacement. How much further does the second projectile travel, in meters, compared to the first?

Explanation / Answer

Remember that range in projectile motion is given by:

Range = V0x*T

V0x = Initial Horizontal Velocity

T = total time period of motion

For first projectile:

Using 2nd kinematic equation

h = V0y*T + 0.5*ay*T^2

V0y = +V0*sin theta = 14*sin 29 deg = 6.79 m/sec

ay = -g = -9.8 m/sec^2

h = -17 m

So,

-17 = 6.79*T - 0.5*9.8*T^2

4.9*T^2 - 6.79*T - 17 = 0

Solving above quadratic equation

T = [6.79 +/- sqrt (6.79^2 + 4*4.9*17)]/(2*4.9)

T = 2.68 sec (by taking +ve sign)

Now

V0x = 14*cos 29 deg = 12.24 m/sec

So,

R1 = V0x*T = 12.24*2.68 = 32.80 m

For 2nd projectile:

Using same process, but this time V0y will be negative as angle is below the horizontal

Using 2nd kinematic equation

h = V0y*T + 0.5*ay*T^2

V0y = -V0*sin theta = -14*sin 29 deg = -6.79 m/sec

ay = -g = -9.8 m/sec^2

h = -17 m

So,

-17 = -6.79*T - 0.5*9.8*T^2

4.9*T^2 + 6.79*T - 17 = 0

Solving above quadratic equation

T = [-6.79 +/- sqrt (6.79^2 + 4*4.9*17)]/(2*4.9)

T = +1.294 sec (by taking +ve sign)

Now

V0x = 14*cos 29 deg = 12.24 m/sec

So,

R2 = V0x*T = 12.24*1.294 = 15.84 m

So compared to first projectile 2nd projectile will land behind

that distance will be = 15.84 - 32.80 = -16.96 m

Check that if negative sign is required for answer or not.

Please Upvote.

Comment below if you have any doubt.

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