A stunt performer is to run and dive off a tall platform and land in a net in the back of a truck below. Originally the truck is directly under the platform, it starts forward with a constant accelera

Question

A stunt performer is to run and dive off a tall platform and land in a net in the back of a truck below. Originally the truck is directly under the platform, it starts forward with a constant acceleration a at the same instant the performer leaves the platform. If the platform is H above the net in the truck, then the horizontal velocity $u$ that the performer must have as he leaves the platform is -

TestHub · Accepted Answer

To land in the truck, the performer and the truck must cover the same horizontal distance in the same time.

Time of fall:

The vertical motion of the performer is under gravity.

$H = \frac{1}{2} g t^{2}$

$t = \frac{2 H}{g}$

Horizontal distance covered by performer:

$x_{p} = u t$

Horizontal distance covered by truck:

$x_{t} = \frac{1}{2} a t^{2}$

Equating horizontal distances:

$u t = \frac{1}{2} a t^{2}$

$u = \frac{1}{2} a t = \frac{1}{2} a \frac{2 H}{g} = a \frac{H}{2 g}$

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JEE Main 2027 (Class 12)