Question

If 4 seconds be the time in which a projectile reaches a point p of its path and 5 seconds the time from p till it reaches the horizontal plane passing through the point of projection. The height of p above the horizontal plane will be

Answer 1

Answer: $H = 98 meter$

Given:  

Time to reach the point p = 4 sec

Time to reach the horizontal plane = 5 sec

Total time traveled by projectile to reach the plane = 4 + 5 = 9 seconds

We know that, time of flight,

                     $T=\frac { 2\quad \times \quad usina }{ g }$

(since u = u sina for vertical directions)

                     $9=\frac { 2\quad \times \quad usina }{ 9.8 }$

Using $u\sin { a } =\quad \frac { 9\quad \times \quad 9.8 }{ 2 } \quad =\quad 44.1$ to find the height of the point P.

We know that,

                     $s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } at^{ 2 }$

                     $H\quad =\quad \left( u\sin { a }\right) t\quad -\quad \frac { 1 }{ 2 } gt^{ 2 }$ (since a = gravitational force and motion towards downwards)

                     $H\quad =\quad \left( 44.1 \right) t\quad -\quad \frac { 1 }{ 2 } \quad \times \quad 9.8\quad \times \quad t^{ 2 }$

Time to reach point p is 4 sec.

                     $H\quad =\quad \left( 44.1 \right) \left( 4 \right) \quad -\quad \frac { 1 }{ 2 } \times \quad 9.8\quad \times \quad (4)^{ 2 }$

                     $H\quad =\quad 176.4\quad -\quad \left( 4.9\quad \times \quad 16 \right) \quad =\quad 176.4\quad -\quad 78.4\quad =\quad 98\quad meter$