几条好难的英文数学题

ThequestionsareonthetopicofprojectilemotioninPhysics.Thehorizontalmotionisinauniformvelocity.Theverticalmotionisinauniforma=g=-10m/s²(Takeupwardquantityaspositive.)=====1*Verticalmotion:u=35sin60°m/s

a=-10m/s²

v=0m/s

t=?sv=u+at0=(35sin60°)+(-10)tTimetaken

t=3.0s=====2*Verticalmotion:u=18sin40°m/s

a=-10m/s²

t=2s

s=?ms=ut+(1/2)at²s=(18sin40°)(2)+(1/2)(-10)(2)²s=3.1mHeightabovethegroundafter2s=5+3.1=8.1m=====3*a)Verticalmotion:u=0m/s

a=-10m/s²

t=2.5s

s=?ms=ut+(1/2)at²s=0+(1/2)(-10)(2.5)²=-31.3mHeight=31.3mb)Horizontalmoton:v=32m/s

t=2.5s

s=?mDistance=vt=(32)(2.5)=80m=====4*a)Verticalmotion(tothehighestpoint):u=150sin10°m/s

v=0m/s

a=-10m/s²

t=?sv=u+at0=150sin10°+(-10)tTimetaken

t=2.6sb)Verticalmotion(tohittheground):u=150sin10°m/s

s=0m

a=-10m/s²

t=?ss=ut+(1/2)at²0=(150sin10°)t+(1/2)(-10)(t²)5t(t-30sin10°)=0t=0(rejected)ort=30sin10°=5.2sHorizontalmotionuntiltheparticlehitstheground:v=150cos10°m/s

t=5.2sRangeoftheprojectile

s=vt=(150cos10°)(5.2)=768m=====5*a)Verticalmotion(tothegreatestpoint):u=20sin45°m/s

v=0m/s

a=-10m/s²

s=?mv²=u²+2as0=(20sin45°)²+2(-10)sGreatestheight

s=10mb)Verticalmotion(tohittheground):u=20sin45°m/s

s=0m

a=-10m/s²

t=?ss=ut+(1/2)at²0=(20sin45°)t+(1/2)(-10)(t²)5t(t-4sin45°)=0t=0(rejected)ort=4sin45°=2.8sHorizontalmotion(tohittheground):v=20cos45°m/s

t=2.8sDistanceOX

s=vt=(20cos45°)(2.8)=40m

参考:andrew