The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that
a ball thrown with a speed of 40 ms can go without hitting the ceiling of the hall. {Don't use calculator for calculating angle. I need steps so the answer is 150.5m} Class 11
Answers
Answered by
2
Since the ceiling of hall is 25 m high therefore the max. height attained by the ball will be 25m.
i.e. H = 25m, u = 40 m/s [given]
we know, H = u2sin2θ / 2g
or, sin2θ = 2gH / u2 = 2 x 9.8 x 25 / ( 40 )2 = 0.30625
or, sin θ = 0.5534
or, θ = 33.6 o
Therefore, R = u2sinθ / g = (40)2 sin 67.2o /9.8
= 150.5 m
Hope it helps..!!
i.e. H = 25m, u = 40 m/s [given]
we know, H = u2sin2θ / 2g
or, sin2θ = 2gH / u2 = 2 x 9.8 x 25 / ( 40 )2 = 0.30625
or, sin θ = 0.5534
or, θ = 33.6 o
Therefore, R = u2sinθ / g = (40)2 sin 67.2o /9.8
= 150.5 m
Hope it helps..!!
Answered by
7
ANSWER
REFER TO THE ATTACHMENT.....
Attachments:
Similar questions