From the top of a building 19.6 m high, a ball is projected horizontally. After how long does it strike the ground? If the line joining the point of projection to the point where it hits the ground makes an anglle 45 degree with the horizontal , What is the initial velocity of the ball?
Answers
Answered by
142
T = (2h/g)
= (2 x 19.6 m/(9.8 m/s2))
= 2 s.
U = 19.6/2
= 9.8 m/s^2
Hope this helps!
= (2 x 19.6 m/(9.8 m/s2))
= 2 s.
U = 19.6/2
= 9.8 m/s^2
Hope this helps!
Answered by
74
"The height of the building is 19.6 m. When the ball hits the ground, it makes the angle of 45 degrees.
Then the initial velocity of the ball is calculated below.
It is a known fact that T = (2h/g)
T = (2 x 19.6 m/(9.8 m/s2))
The value of T is 2 s.
U = 19.6/2
The value of U is 9.8 m/s^2
"
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