hi guys ,, got this problem about projectile ,,
A ball is thrown from the top of a building with a velocity of 40 m/s at an angle of 53
with the horizontal. After 2 s, it is seen to be at height of 84 m above the ground.
a) Find the height of the building
b) At which other time will the ball again be at a height of 84 m?
c) If the ball hits a wall at a height of 52 m above the ground, what is the distance of the
wall from the building?
d) Find the magnitude and direction of the velocity of the ball when it reaches the wall
The Attempt at a Solution
i tried to get the height of the building by 84+h(b)=40sin(53)*2-9.8*4 ,, and I got h(b)=59.31m ,, anyway ,, the way is right ?? ,, and about (b) i tried to get the second time ,, but i get it negative (59.31+84=40sin(53)x-9.8*x^2 = 9.8x^2-40sin(53)x-143.31 and i get x=5.7 , x=-2.5 and i think it's wrong because none of them equal 2 ,, and one of them should equal 2 ,, any ideas what did I do wrong here ?
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