Question

A motor cyclist is trying to jump across a path as shown in fig. by driving horizontally off a cliff A

at a speed of 5m/s. Ignore air resistance and take g =10ms^2. The speed with which he touches peak B is???​

Answer 1

Question:-

Q) A motor cyclist is trying to jump across a path as shown in fig. by driving horizontally off a cliff A  at a speed of 5m/s. Ignore air resistance and take g =10ms^2. The speed with which he touches peak B is??

Answer:-

Speed in horizontal direction remains constant during whole journey because there us no acceleration in this direction .So,

$V_n = 5ms^{-1}$

In Vertical direction, loss in gravitation potential energy = gain in KE, i.e.,

∴ P.E = K.E

mgh = $\frac{1}{2}$ m$v_{v}^{2}$

$v_{v}^{2} = 2gh = 2 * 10 *(70-60)\\ = 200$

therefore :-

the speed with which touches  the  peak B is:-

⇒ $v = \sqrt{v_n^2 + v_v^2}$

⇒ $v = \sqrt{25 + 200}$

⇒ $v = \sqrt{225}$

⇒ $v = 15 ms^-1$

thank  you............