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

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
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

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
(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

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子弹是从悬崖之上125米65的初始速度的地 面边缘的一个镜头。0m／s在水平37°角 （1）确定的弹丸采取点P在地面的时间（2 ）确定的范围内的X弹体的测定从悬崖的底部 在瞬间就在弹丸撞击点 （3）的水平和垂直结构 （4）的速度的大小（5）通过与速度矢量的 角 （6）发现的最大高度以上的悬崖顶上的R进 入由弹...

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翻译了一下，希望对你有帮助
子弹是从悬崖之上125米65的初始速度的地 面边缘的一个镜头。0m／s在水平37°角 （1）确定的弹丸采取点P在地面的时间（2 ）确定的范围内的X弹体的测定从悬崖的底部 在瞬间就在弹丸撞击点 （3）的水平和垂直结构 （4）的速度的大小（5）通过与速度矢量的 角 （6）发现的最大高度以上的悬崖顶上的R进 入由弹

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It's easy to calculate, first determine the vertical velocity is 65*sin37=39m/s, and horizontal velocity is 65*cos37=52m/s, and the projectile can reach the hight is 138m,(1/2mV~2=mgh), then find the ...

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It's easy to calculate, first determine the vertical velocity is 65*sin37=39m/s, and horizontal velocity is 65*cos37=52m/s, and the projectile can reach the hight is 138m,(1/2mV~2=mgh), then find the total time it hit the P point is 7.3s(t= 2h/g~1/2),x=52*7.3=381m.
Just for your instace,please ingore the miscalculation if possible.

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