A man is sitting on the shore of a river. He is in the line of 1m long boat and is 5.5m away form the centre of the boat . He wishes to throw an apple only with a speed of 10m/s , find the minimum and maximum angles of projection for successful shots.

Ayush Gupta, 6 years ago

Grade:11

3 Answers

swarup kumar majee

49 Points

6 years ago

As,per the question the boat is 1m long

and he is 5.5m away from centre of the boat

so,the distance between the one end to the centre of boat is 0.5m as the total lenght of the boat is 1m

and the distance between the other end to the centre of boat is also 0.5m

so,the nearest distance for successful shot is 5.5m-0.5m=5m

and the farthest distance for successful shot is 5.5m+0.5m=6m

now,let us take nearest distance as horizontal range=5m

u=10m/s

and as we know the formula that horizontal range=u²sin2 $\theta$ $\div$ g let,horizontal range be R

so,sin2 $\theta$ =R*g $\div$ u²=1/2

then 2 $\theta$ =30

then^, $\theta$ =15 so,minimum angle of projection is 15

now the maximum angle of projection is,

sin2 $\theta$ =maximum distance*g $\div$ u²=6/10

then 2 $\theta$ =37

then, $\theta$ =18.5

so,the maximum angle of projection for successful shot is 18.5

and the minimum angle of projection for successfull shot is 15

Santosh Kumar Thakur

20 Points

6 years ago

Distance b/w man and centre of the boat = 5.5m

Length of the boat = 1m

So, the distance b/w nearest edge of the boat and man = 5.5m – ½ (1m) = 5m

Then,

Horizontal distance = X(let) = 5 m

Now, let the angle of projection be $\theta$ .

For Horixontal axis

Initial velocity = U = u cos $\theta$

Acceleration = 0

Time = 2 u sin $\theta$ (g)^-1 [From the formula or you can derive by equating the time of vertical axis]

Again,

Horizontal Range = X = u cos $\theta$ { 2 u sin $\theta$ (g)^-1}

=> 5 = 10 cos $\theta$ { 2 (10) sin $\theta$ (10)^-1} [Taking g=10ms^-2]

=> 1 = 2cos $\theta$ sin $\theta$

=> 1 = sin 2 $\theta$ [Since, sin 2 $\theta$ = 2sin $\theta$ cos $\theta$ ]

=> sin 30⁰ = sin 2 $\theta$

So, $\theta$ = 15⁰

Hence, the max angle for successful projection = 90⁰- $\theta$ = 75⁰ [Horizontal range remains same on both angles if they are complementry angles]

and the min angle for successful projection = $\theta$ = 15⁰

Gurmeet

46 Points

6 years ago

As,per the question the boat is 1m longand he is 5.5m away from centre of the boatso,the distance between the one end to the centre of boat is 0.5m as the total lenght of the boat is 1mand the distance between the other end to the centre of boat is also 0.5mso,the nearest distance for successful shot is 5.5m-0.5m=5mand the farthest distance for successful shot is 5.5m+0.5m=6mnow,let us take nearest distance as horizontal range=5m u=10m/sand as we know the formula that horizontal range=u2sin2\theta\divg let,horizontal range be Rso,sin2\theta =R*g\divu2=1/2then 2\theta=30\degreethen,\theta=15\degree so,minimum angle of projection is 15\degreenow the maximum angle of projection is,sin2\theta=maximum distance*g\divu2=6/10then 2\theta =37\degreethen,\theta=18.5\degree so,the maximum angle of projection for successful shot is 18.5\degreeand the minimum angle of projection for successfull shot is 15\degree