Physics - Kinematics Question with Solution | TestHub

Under continuous gravitation effect Time period $=\frac{2(25)\sin 37^{\circ }}{g}=3\sec$ and max height $=11.25$ BC and DE are straight line paths . Hence, net time period $=5\sec$

Particle will be at highest point after 2.5 sec

$$\begin{array}{rlrlrl}& {\ell }_1+{\ell }_2+{\ell }_3=\frac{(25{)}^2{\sin }^2\left(37^{\circ }\right)}{g}=60m\frac{(25{)}^2\sin 2\left(37^{\circ }\right)}{g}=60\mathrm{m}& & & & \end{array}$$

$$x=\left(25\cos 37^{\circ }\right)\times (1)=20m,Hence,range=60+2(20)=100\mathrm{m}$$

$$at\mathrm{t}=1\mathrm{s}{V}_{By}=V_{Ay}-g(1)V_{By}=5m/s$$

$$\mathrm{Y}=5(1)=5\mathrm{m},Effectivemax.height=11.25+5=16.25\mathrm{m}$$

$$V_{Avg.}=\frac{\Delta s}{\Delta t}=\frac{100}{5}=20m/s$$