Question

While taking a test in the testing center, someone mistakes the emergency exit door for the test return door and sets off the alarm. One worker tosses her keys, with a velocity of 9 m/s and at an angle of 30.8 degrees relative to the ground, to the worker closest to the ringing door to shut it off. Assuming starting and ending heights are the same (1.2 m), what was the max height the keys were in the air?

Answer 1

Step 1: Find max height achieved by key during projectile motion w.r.t. launching point:

H = V0y^2/(2*g) = (V0*sin )^2/(2*g)

V0 = Launching speed = 9 m/s

= launching angle = 30.8 deg

So,

H = (9*sin 30.8 deg)^2/(2*9.81)

H = 1.08 m

Step 2:

Now given that starting and ending heights are the same at 1.2 m from ground, So

max height the keys were in the air = H_max = H + h

H_max = 1.08 + 1.20

H_max = 2.28 m

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