A ball is tossed from the window of a building with an initial velocity of 8 m s⁻¹ at an angle of 20° below the horizontal. It strikes the ground 3 s later. From what height was the ball thrown? How far from the base of the building does the ball strike the ground?
Answers
Answered by
14
Hii dear,
# Given-
θ = 20°
t = 3 s
u = 8 m/s
# Calculations-
x = utcosθ
x = 8×3×cos20°
x = 24×0.94
x = 22.6 m
y = utsinθ - 1/2 gt^2
y = 8×3×sin20° - 0.5×9.8×9
y = 24×0.34 - 4.9×9
y = 8.16-44.1
y = -35.94 m
Hence, height of the building is 35.94 m and ball strikes at a distance of 22.6 m from building...
# Given-
θ = 20°
t = 3 s
u = 8 m/s
# Calculations-
x = utcosθ
x = 8×3×cos20°
x = 24×0.94
x = 22.6 m
y = utsinθ - 1/2 gt^2
y = 8×3×sin20° - 0.5×9.8×9
y = 24×0.34 - 4.9×9
y = 8.16-44.1
y = -35.94 m
Hence, height of the building is 35.94 m and ball strikes at a distance of 22.6 m from building...
Answered by
21
Given :
speed=u= 8 m/s
t=20
t=3 s
A) Horizontal Distance = = u cos θ=8 x cos 20° x 3
=8x0.939x 3
=22.6m
B) Heght =h= ( usin θ) t +1/2 gt²
=8 sin20° x3 +1/2 x9.8 x9
=8.208+44.1
=52.31m
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