Question #258926

An object is launched at a velocity of 25 m/s in a direction making an angle of

30° upward with the horizontal.

a. What is the maximum height reached by the object?

b. What is the total flight time (between launch and touching the ground) of

the object?

c. What is the horizontal range (maximum x above the ground) of the object?

d. What is the magnitude of the velocity of the object just before it hits the

ground?


1
Expert's answer
2021-10-31T18:10:14-0400

Given:

v0=25m/sv_0=25\:\rm m/s

g=9.8m/s2g=9.8\:\rm m/s^2

θ=30\theta=30^{\circ}


(a) The maximum height reached by the object


hmax=v02sin2θ2g=252sin23029.8=8.0mh_{\max}=\frac{v_0^2\sin^2\theta}{2g}=\frac{25^2\sin^230^{\circ}}{2*9.8}=8.0\:\rm m

(d) Time of flight


t=2v0sinθg=225sin309.8=2.6st=\frac{2v_0\sin\theta}{g}=\frac{2*25\sin30^{\circ}}{9.8}=2.6\:\rm s

(e) Range


hmax=v02sin2θg=252sin609.8=48mh_{\max}=\frac{v_0^2\sin2\theta}{g}=\frac{25^2\sin60^{\circ}}{9.8}=48\:\rm m


(d) The final speed is equal to initial speed, so

vf=25m/sv_f=25\:\rm m/s


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