Question

The particle is projected with a velocity of 30 m/s at an angle 40° above a horizontal plane. Find
1) the time for which the particle is in the air
2) the horizontal distance it travels

Answer 1

Answer on Question 49941, Physics, Mechanics — Kinematics — Dynamics The particle is projected with a velocity of 30 m/s at an angle 40 above a horizontal plane. Find 1) the time for which the particle is in the air 2) the horizontal distance it travels Solution Vertical initial speed is

$v_{y0}=v_{0}\sin\alpha=30\cdot\sin 40{}^{\circ}\approx 19.3\,m/s$

Let us find time for which the particle is in the air. It is twice of the tame it takes particle to reach highest point. Hence

$t_{t}=2t_{h}=2{v\over g}=2{19.3\over 9.8}\approx 3.9\,s$

Now we can find horizontal distance. It is equal to total time $t_{t}$ times horizontal velocity

$l=v_{x0}t_{t}=v_{0}\cos\alpha\cdot t_{t}=30\cdot\sin 40{}^{\circ}\cdot 3.9\approx 90\,m$

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