A body is projected upward at an angle of 30 with the horizontal at an initial speed of 200. In how many seconds will it reach the ground? How far from the point of projection will it strike??
Answers
Explanation:
The component of velocity in the vertical-upward direction is,
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/s
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t sec
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t secThe acceleration due to gravity is, g=10m/s
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t secThe acceleration due to gravity is, g=10m/s 2, in the vertical-downward direction.
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t secThe acceleration due to gravity is, g=10m/s 2, in the vertical-downward direction.and the distance traveled is, h=70m, in the vertical-downward direction.
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t secThe acceleration due to gravity is, g=10m/s 2, in the vertical-downward direction.and the distance traveled is, h=70m, in the vertical-downward direction.So here, h=−vt+ 21gt 2
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t secThe acceleration due to gravity is, g=10m/s 2, in the vertical-downward direction.and the distance traveled is, h=70m, in the vertical-downward direction.So here, h=−vt+ 21gt 2 ⟹70=−25t+(0.5×10×t2)
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t secThe acceleration due to gravity is, g=10m/s 2, in the vertical-downward direction.and the distance traveled is, h=70m, in the vertical-downward direction.So here, h=−vt+ 21gt 2 ⟹70=−25t+(0.5×10×t2)⟹t^2−5t−14=0
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t secThe acceleration due to gravity is, g=10m/s 2, in the vertical-downward direction.and the distance traveled is, h=70m, in the vertical-downward direction.So here, h=−vt+ 21gt 2 ⟹70=−25t+(0.5×10×t2)⟹t^2−5t−14=0Solving above quadratic equation, we get t=−2,7
The component of velocity in the vertical-upward direction is,v=50sin(300)=50×0.5=25m/sLet the time taken to reach ground from initial position be t secThe acceleration due to gravity is, g=10m/s 2, in the vertical-downward direction.and the distance traveled is, h=70m, in the vertical-downward direction.So here, h=−vt+ 21gt 2 ⟹70=−25t+(0.5×10×t2)⟹t^2−5t−14=0Solving above quadratic equation, we get t=−2,7As time is always positive, t=7s