An object is thrown upwards at an angle of 37 degrees to the horizontal with a speed of 10 ms^-1 from the top of a building 20m high.
a)Find the velocity of the body when it reaches the ground.
ANS: 22.2 ms-1 directed 68.9 degrees below the horizontal
I know how to get the velocity, but i dont get how you get the angle.
Thanks
a)Find the velocity of the body when it reaches the ground.
ANS: 22.2 ms-1 directed 68.9 degrees below the horizontal
I know how to get the velocity, but i dont get how you get the angle.
Thanks
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a) the object is gonna go upwards and then it will fall down till it reaches the floor. when the object is falling and is 20 m from the floor, it's vertical velocity will be the same as it was in the throwing. therefore:
v = v0 + at
v² = v0² + 2ax =
v² = (10*sen37º)² + 2*10*20
(considering that the gravity is 10m/s²)
v = 20.9m/s (that's only the vertical component of the velocity)
the horizontal component of the velocity is V = 10*cos37º
therefore the net velocity is:
V'² = V² + v²
V' = 22.4 m/s (it's a little different from the answer cause I used g = 10m/s² instead of g = 9.8m/s².
to find the angle we need to do as follows:
sen(theta) = 20.9/22.4
therefore theta = arcsen(20.9/22.4) = 68.9º
;D
v = v0 + at
v² = v0² + 2ax =
v² = (10*sen37º)² + 2*10*20
(considering that the gravity is 10m/s²)
v = 20.9m/s (that's only the vertical component of the velocity)
the horizontal component of the velocity is V = 10*cos37º
therefore the net velocity is:
V'² = V² + v²
V' = 22.4 m/s (it's a little different from the answer cause I used g = 10m/s² instead of g = 9.8m/s².
to find the angle we need to do as follows:
sen(theta) = 20.9/22.4
therefore theta = arcsen(20.9/22.4) = 68.9º
;D