Question

A projectile is fired at an angle of 45 with the horizontal. elevation angle of the projectile at its highest point as seen from the point of projection is

a) tan-1(31/2/2)

b) 45

c)60

d)tan-11/2

Answer 1

let say the projectile is projected upwards with speed "v" at an angle of 45 degree

now when it will reach at the highest point its height is given as

$H = \frac{v^2 sin^245}{2g}$

also the horizontal distance of the object is half of the range so it is given as

$X = \frac{v^2 sin(2*45)}{2g}$

now the angle of elevation of that point is given as

$tan\theta = \frac{H}{X}$

$tan\theta = \frac{\frac{v^2sin^245}{2g}}{\frac{v^2 sin(2*45)}{2g}}$

$tan\theta = \frac{\frac{1}{2}}{1}$

$\theta = tan^{-1}\frac{1}{2}$

so correct answer is option D

Answer 2

Use the formula R=4Hmax/tanQ

if you will solve it placing Q=45° then the equation will be R=4Hmax(consider it as equation 1)

Now make a right angle triangle considering line AB (from the point of projection to Hmax) and line AC (from the point of projection to R/2 i.e; half range.

Now you have perpendicular Hmax and Base R/2.

Write tanQ for this triangle which will be tanQ=Hmax/(R/2)

Place the value of R from equation 1

You will get tanQ=Hmax/(4Hmax/2)

After solving it you will get the tanQ which will be tanQ=1/2 i.e; option (d)