Question

For a projectile on the horizontal surface of a planet $y=8t-5t^{2} \, m$ shows height and $x=8\,t \, m$ shows the horizontal distance, then the velocity with which the projectile is projected is

• A. 8 m/s
• B. 6 m/s
• C. 10 m/s
• D. 100 m/s

Answer 1

Answer: (C)

10 m/s

Answer 2

Given, horizontal distance $x=6 t \,m$ vertical distance $y=8 t-5 t^{2} m$ $\therefore$ Horizontal velocity $v_{x}=\frac{d x}{d t}=6\, m / s$ Vertical velocity $v_{y}=\frac{d y}{d t}=8-10 t \,m / s$ at $t=0$, $v_{y}=8\, m / s$ Therefore, resultant velocity $v=\sqrt{v_{x}^{2}+v_{y}^{2}}$ $=\sqrt{6^{2}+8^{2}}=10 \,m / s$