Question #269887

An object is launched at a velocity of 20m/s in a direction making an angle of 25° in a 50 meter elevation. How long is the object in the air? What is the range of the object?

1
Expert's answer
2021-11-22T10:12:38-0500

1. vy=v0sin25°gt1t1=v0sin25°/g=20sin25°/9.8=0.86 (s)1. \ v_y=v_0\sin25°-gt_1\to t_1=v_0\sin25°/g=20\cdot\sin25°/9.8=0.86\ (s)


H=50+v02sin225°/(2g)=50+202sin225°/(29.8)=53.65 (m)H=50+v_0^2\sin^225°/(2g)=50+20^2\cdot\sin^225°/(2\cdot9.8)=53.65\ (m)


H=gt22/2t2=2H/g=253.65/9.8=3.3 (s)H=gt_2^2/2\to t_2=\sqrt{2H/g}=\sqrt{2\cdot 53.65/9.8}=3.3\ (s)


t=t1+t2=0.86+3.34.2 (s)t=t_1+t_2=0.86+3.3\approx4.2\ (s) . Answer


2. x=vxt=v0cos25°t=20cos25°4.276.1 (m)2.\ x=v_xt=v_0\cos25°\cdot t=20\cdot\cos25°\cdot4.2\approx76.1\ (m) . Answer

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