A ball is dropped frm the top of a tower 44.1m high. 2 seconds later another ball iB is htrown downwardsdirection from the same point,both the balls reach the ground at the same time. find the speed of ball B . G = 9.8 M/S^2
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Here in case of BALL A:
distance=44.1 m
initial velocity = 0m/s
g=9.8 m/s²
acc. to second law of motion,
s = ut + 0.5*a*t²
44.1 = 0 + 4.9*t²
√(44.1/4.9) = t
t = 3 seconds
in BALL B:
ball b is thrown after 2 seconds and they reach the ground at the same time
therefore, time for BALL B = 3-2 = 1 second
initial velocity = ?
distance = 44.1 m
s = ut + 0.5*g*t²
44.1 = u + 4.9
u = 44.1-4.9
u = 39.2 m/s
Therefore, BALL B is thrown vertically down with an initial velocity of 39.2 m/s
Hope this Helps :)
distance=44.1 m
initial velocity = 0m/s
g=9.8 m/s²
acc. to second law of motion,
s = ut + 0.5*a*t²
44.1 = 0 + 4.9*t²
√(44.1/4.9) = t
t = 3 seconds
in BALL B:
ball b is thrown after 2 seconds and they reach the ground at the same time
therefore, time for BALL B = 3-2 = 1 second
initial velocity = ?
distance = 44.1 m
s = ut + 0.5*g*t²
44.1 = u + 4.9
u = 44.1-4.9
u = 39.2 m/s
Therefore, BALL B is thrown vertically down with an initial velocity of 39.2 m/s
Hope this Helps :)
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