A ball is thrown with a velocity whose horizontal component is 12 m s⁻¹ from a vertical pole 15m high in such a way that it just clears the wall of height 18.75m. At what time (in sec) will it reach the ground? (g = 10m s⁻²)
Physics · Mixed Chapter
JEE Mains

Question

A ball is thrown with a velocity whose horizontal component is 12 m s⁻¹ from a vertical pole 15m high in such a way that it just clears the wall of height 18.75m. At what time (in sec) will it reach the ground? (g = 10m s⁻²)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The wall in the figure is 6 m horizontally from the 15 m launch pole, and the horizontal speed is 12 m/s. The projectile reaches the wall in t₁=6/12=0.5 s. Step 2: The projectile just clears its top at 18.75 m, rising 3.75 m by t=0.5 s. Vertical equation 3.75=uᵧ(0.5)−½(10)(0.5)² gives 3.75=0.5uᵧ−1.25, so uᵧ=10 m/s. Step 3: For ground impact from launch height 15 m, solve 0=15+10t−5t² ⇒ t²−2t−3=0. Step 4: Positive root t=3 s. Answer: 3.