Question

An Olympic diver springs off of a high dive that is 3 m above the surface of the water. When she lands in the water she is traveling at a speed of 8.90 m/s at an angle of 75.0° with respect to the horizontal. What was her take off speed and direction, up from the horizontal?

Answer 1

Answer:$u=4.51 m/s\ at\ angle\ of\ \theta =59.34^{\circ}$

Explanation:

Given

height of building h=3 m

Landing velocity of diver $v=8.90 m/s$ at an angle of $75^{\circ}$

Let u be the initial velocity of diver at an angle of \theta with horizontal

Since there is no acceleration in horizontal direction therefore horizontal component of velocity will remain same

$u\cos \theta =8.9\cos (75)$        ---- -----1

Considering Vertical motion

$v^2-u^2=2as$

here $v=8.9\sin (75)$

$u=u\sin \theta$

$s=3 m$

$a=9.8 m/s^2$

$(8.9\sin (75))^2-(u\sin \theta )^2=2\times 9.8\times 3$

$u\sin \theta =\sqrt{(8.9\sin 75)^2-(2\cdot 9.8\cdot 3)}$

$u\sin \theta =3.886$         ----------------2

Divide 2 and 1 we get

$\tan \theta =\frac{3.886}{8.9\cos (75)}$

$\tan \theta =1.687$

$\theta =59.34^{\circ}$

Thus $u\cos (59.34)=8.9\cos (75)$

$u=4.51 m/s$