A ball is projected upwards from top of tower with velocity 50 m/s making an angle of 30 degree wth horizontal the height of tower is 70 m after how much time from instant of throwing will ball reach ground
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Let throwing point be (0,0). Horizontal be x-direction and vertical be y-direction
Initial velocity = (50cos30) i + (50sin30) j,
a = -10 j
s = ut + 0.5*a*t2,s = (50*cos30*t) i + (50*sin30*t – 5*t2) j
it will reach the ground when y-coordinate of ball is -70,
So, (50sin30*t – 5t2) = -70
or, 25*t-5*t2=-70,
t2-5t-14=0(t-7)(t+2)=0
=> t=7
Initial velocity = (50cos30) i + (50sin30) j,
a = -10 j
s = ut + 0.5*a*t2,s = (50*cos30*t) i + (50*sin30*t – 5*t2) j
it will reach the ground when y-coordinate of ball is -70,
So, (50sin30*t – 5t2) = -70
or, 25*t-5*t2=-70,
t2-5t-14=0(t-7)(t+2)=0
=> t=7
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44
Answer:
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