Question

At a height of $15m$ from ground velocity of a projectile $\overrightarrow{v}=(10\hat{i}+10\hat{j})$ and ($g=10m{s}^{-2}$ )

• A. particle was projected at an angle of ${45}^{\circ }$ with horizontal
• B. time of flight of projectile is $4s$
• C. horizontal range of projectile is $100m$
• D. maximum height of projectile form ground is $20m$

Answer 1

Answer: (B), (D)

time of flight of projectile is $4s$
maximum height of projectile form ground is $20m$

Using kinematics IIIrd equation

$v^2-u^2=2gS$

${10}^2-u_y^2=-2g\times 15$

$100+300=u_y^2$

$u_y=20m/s$

$u_x=10m/s$

From above values of $u_x$ and $u_y$ it is clear that the angle of projection is not ${45}^o$

$T=\frac{2u_y}{g}=4s$

$Range=u_x\times T=10\times 4=40m$

maximum height of projectile $h_{max}=\frac{u_y^2}{2g}=\frac{{20}^2}{2\times 10}=20m$