Question #338229

A ball is thrown vertically upward with a speed of 7.1 m/s. Find:




(9 points)




a. Maximum height reached the ball




b. Its time of flight




c. Velocity the ball would return to its starting point

1
Expert's answer
2022-05-09T09:45:05-0400
v(t)=v0gtv(t)=v_0-gt

h(t)=h0+v0tgt22h(t)=h_0+v_0t-\dfrac{gt^2}{2}

a. The ball reaches a maximum height at


t1=v0gt_1=\dfrac{v_0}{g}

hmax=h0+v022gh_{max}=h_0+\dfrac{v_0^2}{2g}

Given h0=0m,v0=7.1m/s,g=9.81m/s2h_0=0m, v_0=7.1m/s, g=9.81m/s^2


hmax=0m+(7.1m/s)22(9.81m/s2)=2.569mh_{max}=0m+\dfrac{(7.1m/s)^2}{2(9.81m/s^2)}=2.569m

b.


tflight=2t1=2v0gt_{flight}=2t_1=\dfrac{2v_0}{g}

tflight=2(7.1m/s)9.81m/s2=1.4475st_{flight}=\dfrac{2(7.1m/s)}{9.81m/s^2}=1.4475s

c.


vreturn=v0gtflight=v0v_{return}=v_0-gt_{flight}=-v_0

vreturn=7.1m/sv_{return}=-7.1m/s


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