Question

A boy stands at 39.2 m from a building and throws a ball which just passes through a window 19.6m above the ground. Calculate the velocity of projection of the ball

Answer 1

HEY THERE

GIVEN :-

Maximum height  or H = 19.6 m

Range , R = 39.2  + 39.2     = 78.4 m

P.S. :- Do remember a relation between height and range '

H / R = 1/4 tan∅

⇒ 19.6 × 4 ÷ 78.4 = tan ∅

⇒ ∅  = 45 °

now using range formula that is ,

R = u² sin2Ф÷g

⇒ u² = 78.4 × 9.8

⇒u   = √78.4×9.8

⇒u    = 27.7  m/s

HENCE THE VELOCITY OF THE PROJECTION OF THE BALL IS  27.7 m/s

Answer 2

Answer:

GIVEN :-

Maximum height  or H = 19.6 m

Range , R = 39.2  + 39.2     = 78.4 m

H / R = 1/4 tan∅

⇒ 19.6 × 4 ÷ 78.4 = tan ∅

⇒ ∅  = 45 °

R = u² sin2Ф÷g

⇒ u² = 78.4 × 9.8

⇒u   = √78.4×9.8

⇒u    = 27.7  m/s

HENCE THE VELOCITY OF THE PROJECTION OF THE BALL IS  27.7 m/s