a stone is thrown from a height of 4 metre high building at its highest flight the stone is just cross the top of 10 m high tree from ground .trace the path of projectile if the horizontal distance between building and tree is 5 m.find the distance of the point from the building where the stone fall on the ground
Answers
Given: stone is thrown from a height of 4 metre high building at its highest flight the stone is just cross the top of 10 m high tree from ground
horizontal distance between building and tree is 5 m
To Find : distance of the point from the building where the stone fall on the ground
Solution:
Horizontal Distance crossed = 5 m ( at top of tree)
Vertical Distance crossed = 6 m
Speed = V at angle α
V Cosα T = 5 ( Horizontal Distance ) till tree ( at Peak ) )
Using V² - U² = 2aS
a = - g = - 10 S = 10 - 4 = 6 , vertical velocity at top = 0
0² - V²Sin²α = 2(-10)(6)
=> V²Sin²α = 120
=> V Sinα = √120
Using V = U + at
0 = V Sinα - 10T => T = V Sinα/10
=> T = √120 /10
V Cosα T = 5
=> V Cosα(√120 /10) = 5
=> V Cosα = 50/√120
=> V Cosα =25/√30
V Sinα = √120
V Cosα = 25/√30
=> V = 65/√30
=> tanα = 60/25 = 12/5 => α = 67.38°
S = Ut + (1/2)at² ( time taken from peak to ground)
=> 10 = (1/2)(10)t²
=> t = √2
Horizontal Distance covered after tree = V Cosα * t
=(25/√30) * √2
= 25/√15
= 6.45 m
5 + 6.45 = 11.45 m is distance of the point from the building where the stone fall on the ground
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