Question #5c760

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Answer 1 · 2018-01-25T21:59:42

(a)

$sf(v_0=8.9color(white)(x)"m/s")$

(b)

She will just make the bank at 10 m.

(a)

The expression for range is given by:

$sf(d=(v_0^2sin2theta)/g)$

$:.$ $sf(v_0^2=(dg)/(sin2theta)=(8.0xx9.81)/(sin90)=78.48)$

$sf(v_0=sqrt(78.48)=8.86color(white)(x)"m/s")$

(b)

MFDocs

The key to this is to find the time of flight, which is common to both the vertical and horizontal components of motion:

Using the equation of motion:

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

If we set the launch point as the origin this becomes:

$sf(-2.5=vsinthetat-1/2"g"t^2)$

$sf(-2.5=8.86sin45t-1/2xx9.81t^2)$

$sf(-2.5=8.86xx0.707t-4.905t^2)$

$sf(4.905t^2-6.26t-2.5=0)$

We can use the quadratic formula to solve this. Ignoring the -ve root this gives:

$sf(t=1.596color(white)(x)s)$

The horizontal component of velocity is constant and is equal to $sf(v_0cos45)$ so we can write:

$sf(d=vxxt=8.86cos45xx1.596=10color(white)(x)m)$

She just makes it.