A lead shot is projected from tha ground level with a velocity u at an angle \(\theta\) to the horizontal. Given the time, t for the lead shot to reach its ...
Question 1 Report
A lead shot is projected from tha ground level with a velocity u at an angle \(\theta\) to the horizontal. Given the time, t for the lead shot to reach its maximum height as; t = \(\frac{u \sin \theta}{g}\) where "g" is the acceleration of free fall due to gravity, show that the greatest height reached by the body is h\(_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}\)
Answer Details
For the vertical motion of the projectile, the initial vertical velocity is \(u\sin\theta\) and the acceleration due to gravity is \(-g\).
Using
\[ h=(u\sin\theta)t-\frac{1}{2}gt^{2} \]At maximum height,
\[ t=\frac{u\sin\theta}{g} \]Therefore,
\[ h_{\text{max}}=(u\sin\theta)\left(\frac{u\sin\theta}{g}\right) -\frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2 \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{1}{2}g\left(\frac{u^2\sin^2\theta}{g^2}\right) \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{u^2\sin^2\theta}{2g} \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{2g} \]Hence, the greatest height reached is
\[ \boxed{h_{\text{max}}=\frac{u^2\sin^2\theta}{2g}} \]Download The App On Google Playstore
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