# [物理]大學物理題目(求詳解)

1.A basketball player throws the ball with initial velocity Vo at θ=30 above the horizontal to the hoop which is located a horizontal distance L=2.5m and at a height h=1m above the release point. Find the initial velocity Vo?

2.A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 260 m below. If the helicopter is traveling horizontally with a speed of 60 m/s, (a) how far in advance of the recipients (horizontal distance) must the package be dropped? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?

### 1 個解答

• 最佳解答

1.假設重力加速度g=-10m/s2 (定向下為負)

水平速度Vx=Vocos30°=√3/2Vo=0.866Vo

鉛直速度Vy=Vosin30°= 1/2Vo= 0.5Vo

X=VxT ==> 0.866Vo*T = 2.5 ---------(1)

H=VyT+0.5gT^2 => 0.5VoT-0.5*10*T^2=1---(2)

由(1)得T = 2.5/0.866Vo= 2.886/Vo代入(2)

0.5Vo*2.886/Vo-0.5*10*8.33/Vo^2 =1

=>1.443-41.667/Vo^2 =1

=>0.443Vo^2=41.667

=> Vo = 9.7 m/s

2.

(a)先算出落下時間 T

1/2gT^2=260 => T^2=52 => T= √52 s

horizontal distance X=VxT=60*√52 =432.7 m

(b)60*t=400 ==> t= 20/3=6.67 s

Vyt+0.5gt^2=-260

=>6.67Vy-5*6.67^2=-260

=> Vy =-5.67 m/s

=> 5.67 m/s向下

(c) 落地鉛直速度vy= Vy+gt=-5.67-10*6.67=-72.33 m/s

落地水平速度vx=60 m/s

落地速度V=√vx^2+vy^2 =94 m/s

參考資料： Four-Year