Question

A projectile is thrown at an angle 37° from the
vertical. The angle of elevation of the highest point
of the projectile from point of projection is​

Answer 1

Hey Dear,

◆ Answer -

Angle of elevation = 20.64°

● Explaination -

# Given -

θ = 37°

# Solution -

Ratio of H/R is given by -

H/R = tanθ / 4

H/R = tan37° / 4

H/R = 0.7536 / 4

H/R = 0.1884

Angle of inclination of hinghest point is -

tanα = H / (R/2)

tanα = 2 × H/R

tanα = 2 × 0.1884

tanα = 0.3768

Taking tan inverse -

α = arctan(0.3768)

α = 20.64°

Therefore, angle of elevation of highest point will be 20.64°.

Thanks...