Question

A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR. a) At what angleθis therock thrown? b) In terms of its original rangeR, what is the rangeRmaxtherock can attain if it is launched at the same speed but at the optimal anglefor maximum range? c) Would your answer to part a) be different if the rockis thrown with the same speed on a different planet? Explain.

Answer 1

The range of a projectile motion is given by:

`(u_o^2 sin2\theta)/(g)`

where,
`u` is the initial speed of the projectile,
`\theta` is the angle of the projectile and
`g` is the acceleration due to gravity.

The maximum height reached is given by:

`(u_o^2 sin^2\theta)/(2g)`

Part a

It is given that the maximum height reached is equal to the horizontal range. we need to find the angle of the projectile.

Equating the two:

`(u_o^2 sin2\theta)/(g)=(u_o^2 sin^2\theta)/(2g)\\ \Rightarrow 2 sin2\theta =sin^2\theta\\ \Rightarrow 2* 2 sin\theta cos\theta=sin^2 \theta\\ \Rightarrow tan\theta =4\\ \Rightarrow \theta=tan^(-1)4=75.96^o`

Hence, the projectile was thrown at an initial angle of
`\theta=75.96^o`.

Part b

we need to find the angle for which range would be maximum and then write this maximum range in terms of original range.

So, we know that range is given by:

`R= (u_o^2sin2\theta)/(g)`

It would be maximum when
`sin2\theta=1\Rightarrow 2\theta=90^o\Rightarrow \theta=45^o`

Hence,
`R_(max)=(u_o^2)/(g)`

Original range,

`R=(u_o^2sin2*75.96^o)/(g)\\ \Rightarrow R=(u_o^2)/(g)* 0.47\\ \Rightarrow R=R_(max)0.47\\ \Rightarrow R_(max)=2.125R`

Part c:

In the part a, we know that the angle of the projectile is independent of the
`g` i.e. the acceleration due to gravity and this is the only factor that varies with the different planets. Hence, the answer would remain same.