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 Sep 23rd 2008, 01:09 PM #1 Junior Member   Join Date: Sep 2008 Posts: 14 motion a projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m at an angle of 35 with the horizontal. (a) Determine the time taken by the projectile to hit point P at ground level. (b) Determine the distance X of point P from the base of the vertical cliff (c) At the instant just before the projectile hits point P, find the horizonal and the vertical components of its velocity (d) the magnitude of the velocity (e) the angle made by the velocity with the horizonal (f) find the max height above the cliff top reached by the projectile My attempt (a) t1 = vy/g = 3.8 sec t2 = square root of (2 X s / g) = 6.2 for s i did s= 115 + 37 X 3.8 + (1/2)(9.81)3.8^2 s= 185 T = t1 + t2 = 3.8 + 6.2 = 10 (b) distance = t X Vx 10 X 53 = 531 (f) is s = 185 and im completely lost on c , d, e
 Sep 23rd 2008, 08:47 PM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 quote=whazzap88;1471]a projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m/s at an angle of 35 with the horizontal. (a) Determine the time taken by the projectile to hit point P at ground level. (b) Determine the distance X of point P from the base of the vertical cliff (c) At the instant just before the projectile hits point P, find the horizonal and the vertical components of its velocity (d) the magnitude of the velocity (e) the angle made by the velocity with the horizonal (f) find the max height above the cliff top reached by the projectile [/quote] I don't know whether I'm correct or not. a) $\displaystyle s_y=u_yt-gt^2/2$ $\displaystyle -115= 65sin35^ot-9.8t^2/2$ $\displaystyle t= 9.96s$cor to 3 sig fig b) $\displaystyle s_x= u_xt$ $\displaystyle s_x=65cos35^o(9.964068986)$ $\displaystyle s_x=560 m$cor to 3 sig fig c) As there is no acceleration in horizontal direction, therefore x-component of velocity is = $\displaystyle 65cos35^o$ For y- component of velocity, $\displaystyle v_y=u_yt-gt$ $\displaystyle v_y= 65sin35^o-(9.8)(9.964068986)$ $\displaystyle v_y=-60.3m/s$cor. to 3 sig fig d)$\displaystyle v^2=v_x^2+ v_y^2$ $\displaystyle v=80.5m/s$cor. to 3 sig fig e) angle required=$\displaystyle tan^-1(v_y/v_x)$ angle required = $\displaystyle 48.6^o$ f) $\displaystyle v^2-u^2=2gh$ $\displaystyle 0-(65sin35)^2=-2(9.8)h$ $\displaystyle h=70.9m$cor. to 3 sig. fig.
 Sep 23rd 2008, 10:16 PM #3 Junior Member   Join Date: Sep 2008 Posts: 14 for (b) isnt it 531 rather than 560.. is it typo????
Sep 24th 2008, 07:10 AM   #4

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 Originally Posted by whazzap88 a projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m at an angle of 35 with the horizontal. (a) Determine the time taken by the projectile to hit point P at ground level. (b) Determine the distance X of point P from the base of the vertical cliff (c) At the instant just before the projectile hits point P, find the horizonal and the vertical components of its velocity (d) the magnitude of the velocity (e) the angle made by the velocity with the horizonal (f) find the max height above the cliff top reached by the projectile My attempt (a) t1 = vy/g = 3.8 sec t2 = square root of (2 X s / g) = 6.2 for s i did s= 115 + 37 X 3.8 + (1/2)(9.81)3.8^2 s= 185 T = t1 + t2 = 3.8 + 6.2 = 10 (b) distance = t X Vx 10 X 53 = 531 (f) is s = 185 and im completely lost on c , d, e
b) The method is right, but I'm getting 530 m.

The rest looks good.

-Dan
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