DAREDEVIL Physics !

Classical Mechanics Level pending

A daredevil on a motorbike jumps a river 6m wide. He lands on the edge of the far bank which is 2.5m lower than the bank from which he takes off. His minimum horizontal speed , in m/s, at take off is:

This problem is not original

9.8 8.4 10.8 7.0

1 solution

Mark Mottian
Sep 5, 2014

Vertically:

$s = ut + \frac {1}{2}at^2$

$2.5 = 0 + \frac {1}{2}(9.8)t^2$

Upon solving for $t$ we discover there is $0.714s$ to fall the $2.5m$ between one side and another.

Horizontally:

The minimum speed to get across is clearly achieved when $a = 0$ . The time to fall vertically is exactly the same amount of time to fall horizontally.

$s = ut + \frac {1}{2}at^2$

$6 = u(0.714) + \frac {1}{2}(0)(0.714)^2$

Which gives $u = 8.4 m/s$

nice solution

Mardokay Mosazghi - 6 years, 6 months ago

@Mark Mottian "The time to fall vertically is exactly the same amount of time to fall horizontally" can u explain this please ?

Saraswati Sharma - 4 years, 8 months ago

DAREDEVIL Physics !

1 solution

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