The initial velocity v_0 of a hockey puck is 105 mi/h. Determine a) the largest value (less than 45^o) of the angle a for which the puck will enter the net, b) the corresponding time required for the puck to reach the net.

Question

cheekabooy · Accepted Answer

Soluton of your problem:     .

user_27qwe · Accepted Answer

(a) Horizontal motion. (Uniform)  x=0+(νx)0t=(154cos⁡α)t  At the front of the net,x=16 ft  Then 16=(154cos⁡α)t  or tenter=877cos⁡α  Vertical motion. (Uniformly accelerated motion)  y=0+(νy)0t−12gt2  =154(sin⁡α)t−12gt2 g=32.2 ft/s2  At the front of the net,  yfront=(154sin⁡α)tenter−12gtenter2  =(154sin⁡α)(877cos⁡α)−12g(877cos⁡α)2  =16tan⁡α−32g5929cos2⁡α  Now 1cos2α=sec2⁡α=1+tan2⁡α  Then yfront=16tan⁡α−32g5929(1+tan2⁡α)  or tan2⁡α−59292gtan⁡α+(1+592932gyfront)=0  Then tan⁡α=59292g±[(−59292g)2−4(1+592932gyfront)]122  or tan⁡α=59294×32.2±[(−59294×32.2)2−(1+592932×32.2yfront)]12  or tan⁡α=46.0326±[(46.0326)2−(1+5.7541yfront)]12  Now 0&amp;lt;yfront&amp;lt;4 ft so that the positive root will yield values of a&amp;gt;45° for all values of yfront  When the negative root is selected, α increases as yfront is increased. Therefore, for set αmax, set  yfront=yc=4 ft  Then tan⁡α=46.0326±[(46.0326)2−(1+5.7541yfront)]12  or αmax=14.6604∘  (b) We had found tenter=877cos⁡α  =877cos⁡14.6604∘  tenter=0.1074 s

FSP0251031821.jpgFSZ

Answered question

Answer & Explanation

New Questions in Other