A particle is projected vertically up from the top of a tower with velocity `10 m//s`. It reaches the ground in 5s. Find -
(a) Height of tower
(b) Striking velocity of particle at ground
(c ) Distance traversed by particle.
(d) Average speed & average velocity of particle.

Explanation

Time taken to reach B
`t_(AB) = (u)/(g) = (10)/(10) = 1 sec`.
`t_(AC) = 5sec`. (given so `t_(BC) = 5-1 = 4 sec`.
`S_(BC) = (1)/(2)"gt"^(2) =(1)/(2) (10)(4)^(2) = 80` m
`S_(AB) = (u^(2))/(2g) = ((10)^(2))/( 20 ) = 5`m
(a) Height of tower h `= 80 - 5 = 75` m
(b) ` v_c =0 + "gt"_(BC) = 40 m//s`
(c ) Distance travelled =` 5+ 80 = 85` m
(d) Average speed = `(85)/(5) = 17` m/s
Average velocity = `-(75)/(5) = - 15 m//s`