
Answer:
Height as seen by the professor = 38.2 m
Explanation:
Angle of throw = θ = 69°
Velocity of throw = v
X component of velocity = v₁ = v cos 69 = 0.3584 v m/s
Vertical component of the velocity = v₂ = v sin 69 = 0.9336 v m/s
v₂ / v₁ = tan 69 = 2.605
v₂ = 2.605 v₁.
Professor sees as if the x component of velocity =0
v (as seen by professor) + v' = 0
=> v as seen by professor = -v' = -10.5 m/s
This shows that y component of the ball's velocity is 2.605 times its x component of velocity.
with respect to the professor, there is only y component of velocity.
v₂' =v₂ = 2.605 ( -10.5) = 27.4 m/s.
Height as seen by the professor = (27.4)² / 2(9.8) = 38.2 m