How to find horizontal distance in projectile motion?

Question

a spring board diver leaves the board 3 m above the water with a velocity of 2.3 ms-1 at an angle of 110 degrees to the board. at what horizontal distance from the end of the board will the diver enter the water?

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Answer 1 · 2018-03-05T13:59:18

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The key is to separate the motion into its horizontal and vertical components and use the idea that they share the same time of flight.

I assume the angle to the horizontal is the complementary to $sf(110^@=70^@)$

I will set the launch point as the origin and use the convention that "up is +ve".

I will use:

$sf(s=ut+1/2at^2)$

This becomes:

$sf(s=vsinthetat-1/2"g"t^2)$

$:.$ $sf(-3=2.3sin70t=1/2xx9.81xxt^2)$

$sf(-3=2.161t-4.91t^2)$

$sf(4.91t^2-2.161t-3=0)$

Using the quadratic formula we get:

$sf(t=1.03color(white)(x)s)$

The horizontal component of velocity is constant so we get:

$sf(d=vcosthetat)$

$sf(d=2.3xx0.342xx1.03=0.8color(white)(x)m)$