A ball is thrown at an angle of 30 degrees from a building of height 20 m
Answers
Answer:
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Explanation:
The key to problems like this is to treat the horizontal and vertical components of the motion separately and use the fact that they share the same time of flight.
Horizontal Component
The horizontal component of the velocity is constant as it is perpendicular to g.
So we can write:
vcos30=20t
Where t is the time of flight.
∴v×0.866=20t
t=200.866v (1)
Vertical Component
We can use:
s=ut+12at2
This becomes:
5=vsin30t−12gt2
∴5=v×0.5×t−12×9.8×t2
5=v×0.5×t−4.9t2 (2)
We can substitute the value of t from (1) into the first part of (2)⇒
∴5=v×0.5×200.866v−4.9t2
∴5=100.866−4.9t2
4.9t2=11.547−5
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Answer:
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A ball is thrown upward at an angle of 30° with the horizontal and lands on the top edge of a building
that is 20 m away. The top edge is 5m above the throwing point. The initial speed of the ball in
metre/second is (take g=10 m/s2):
(A) U - 40 114+3)
V 13.13
(B) u = 40
V13 do mis
40
(C)u-40, 4+ mis
(D) u = 40
m/s
v 13
13 (4+13)