A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff $' X'$ at the speed $10 m s^{- 1}$ . When he touches peak $Z$ (ignore air resistance), what would be speed? (use $g = 10 m s^{- 2}$ )

Question

MARKS Content Team · Accepted Answer

Correct answer: (A) 30 m s^(-1). Use the concept of conservation of energy in that case the sum of initial kinetic energy and potential energy should be equal to the sum of final kinetic energy and potential energy, ⇒mgh+1/2mv^2=mgh'+1/2mv'^2⇒v'^2=2gh+v^2-2gh' here, h'=60 m, h=100 m and v=10 m s^(-1), ⇒v'^2=2×10×100+10^2-2×10×60⇒v'=30 m s^(-1)

A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff $' X'$ at the speed $10 m s^{- 1}$ . When he touches peak $Z$ (ignore air resistance), what would be speed? (use $g = 10 m s^{- 2}$ )

Solution

Relevant Questions

A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff 'X' at the speed 10 m s-1. When he touches peak Z (ignore air resistance), what would be speed? (use g=10 m s-2)

Solution

Relevant Questions

A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff $' X'$ at the speed $10 m s^{- 1}$ . When he touches peak $Z$ (ignore air resistance), what would be speed? (use $g = 10 m s^{- 2}$ )