Question

A particle is projected at an angle of t to the horizontal, with a velocity of u. if the maximum distant travelled is 20m and the greatest height reached is 10m. Find u and t.

Answer 1

$u_{0x}=u\cdot\cos t$

$u_{0y}=u\cdot\sin t$

$h=u_{0y}t-gt^2/2$

$u_y=u_{0y}-gt\to t=u_{0y}/g$

$h_{max}=u_{0y}t-gt^2/2=u_{0y}^2/g-u_{0y}^2/(2g)=u_{0y}^2/(2g)=$

$=u^2\cdot\sin^2 t/(2g)$

The maximum distant travelled

$l_{max}=u^2\cdot\sin2t/g$ . We get

$h_{max}/l_{max}=\tan t/4\to 10/20=\tan t/4\to \tan t=2$

$t=63.4°$ . Answer

$u=\sqrt{gl_{max}/\sin2t}=\sqrt{9.8\cdot 20/\sin(2\cdot 63.4°)}=15.6(m/s)$ . Answer