A ball is thrown vertically upwards with velocity 10 m/s from a trolley moving horizontally with velocity of 8 m/s.
A ball is thrown vertically upwards with velocity 10 m/s from a trolley moving horizontally with velocity of 8 m/s. A person standing on ground observe its motion. The range observed by him is (g = 10m/s).
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Vertical velocity of ball uy=v1
Horizontal velocity of ball ux=v2
Horizontal range of projectile R=2uxuy/g
⟹ R=
2v1v2/g = 2* 10*8/ 10= 16 m
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Answered
Horizontal velocity of ball is trolley
S = uHt
R = uHt
Sg = ugt - \(\frac12\)agt2
Sy = 0
uyt - \(\frac12\)gt2 = 0
uyt = \(\frac12\)gt2
uy = \(\frac{gt}2\)
t = \(\frac{2u_y}g\)
t = \(\frac{2\times10}{10}\)
t = 2 sec
Range observed by him is
Given uH = 8m/s
R = uHt
R = 8 x 2
R = 16 m
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