搜狗做题网A projectile is a shot from the edge of a cliff 125m above g
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A projectile is a shot from the edge of a cliff 125m above ground level with an initial speed of 65.0m/s at angle of 37.0° with the horizontal
(1)Determine the time taken by the projectile to hit point P at ground level
(2)Determine the range X of the projectile as measured from the base of the cliff
At the instant just before the projectile hits point P,find:
(3)the horizontal and the vertical coponents of its velocity
(4)the magnitude of the velocity
(5)the angle made by the velocity vector with the horizontal
(6)find the maximum height above the cliff top reached by the projectile
(1)Determine the time taken by the projectile to hit point P at ground level
(2)Determine the range X of the projectile as measured from the base of the cliff
At the instant just before the projectile hits point P,find:
(3)the horizontal and the vertical coponents of its velocity
(4)the magnitude of the velocity
(5)the angle made by the velocity vector with the horizontal
(6)find the maximum height above the cliff top reached by the projectile
(1) In vertical direction,we have the equation of motion y=v0t+1/2gt^2,where v0=65sin37.calculate the time when y=-125.
(2) In horizontal direction,the motion is uniform,x=vt where v=65cos37 and t has been abtained in (1)
(3) In vertical direction,we have the equation of velocity v=v0+gt,where t is again obtained in (1) and in horizontal direction,the velocity is uniform.
(4) Pythagorean theorem v^2=vx^2+vy^2
(5) tan a= vy/vx where a is the ange between the velocity and horizon3
(6) Since we have y=v0t+1/2gt^2,find the maximum value of y by either calculating first derivative or rearranging the equation.Dont forget to add 125m to your result
(2) In horizontal direction,the motion is uniform,x=vt where v=65cos37 and t has been abtained in (1)
(3) In vertical direction,we have the equation of velocity v=v0+gt,where t is again obtained in (1) and in horizontal direction,the velocity is uniform.
(4) Pythagorean theorem v^2=vx^2+vy^2
(5) tan a= vy/vx where a is the ange between the velocity and horizon3
(6) Since we have y=v0t+1/2gt^2,find the maximum value of y by either calculating first derivative or rearranging the equation.Dont forget to add 125m to your result
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