Question #288529

1. Ball thrown at 45° with intial velocity of 25m/s.


Find the


A. Initial velocity


B. Final velocity


C. Acceleration (m/s²)


D. Time ( s )


E. Distance ( m )

Expert's answer

Given:

v0=25 m/sv_0=25\:\rm m/s

θ=45∘\theta=45^\circ


(a)

vix=v0cos⁡θ=25cos⁡45∘=17.7 m/sviy=v0sin⁡θ=25sin⁡45∘=17.7 m/sv_{ix}=v_0\cos\theta=25\cos45^\circ=17.7\:{\rm m/s}\\ v_{iy}=v_0\sin\theta=25\sin45^\circ=17.7\:{\rm m/s}

(b)

vfx=v0cos⁡θ=25cos⁡45∘=17.7 m/svfy=−v0sin⁡θ=−25sin⁡45∘=−17.7 m/sv_{fx}=v_0\cos\theta=25\cos45^\circ=17.7\:{\rm m/s}\\ v_{fy}=-v_0\sin\theta=-25\sin45^\circ=-17.7\:{\rm m/s}

(c)

ax=0 m/s2ay=−g=−9.8 m/s2a_{x}=0\:{\rm m/s^2}\\ a_{y}=-g=-9.8\:{\rm m/s^2}

(d)

t=2v0sin⁡θg=2∗25∗sin⁡45∘9.8=3.6 st=\frac{2v_0\sin\theta}{g}=\frac{2*25*\sin45^\circ}{9.8}=3.6\:\rm s

(e)

d=v02sin⁡2θg=252∗sin⁡90∘9.8=63.8 md=\frac{v_0^2\sin2\theta}{g}=\frac{25^2*\sin90^\circ}{9.8}=63.8\:\rm m


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