A little boy is standing at the edge of a steep vertical cliff. he throws a ball at an angle of 40◦ with respect to the horizontal at an initial speed of 20 m/s. at a certain moment, the ball has fallen 5 m below the level at which it was thrown.

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aristocles aristocles · Answer 1 · 2018-09-16T04:19:54+02:00

boy throws the ball with speed 20 m/s at an angle 40 degree

so the components are given as

[tex]v_x = 20cos40 = 15.3 m/s[/tex]

[tex]v_y = 20 sin40 = 12.8 m/s[/tex]

now the time taken by the ball to drop down by 5 m is given as

[tex]y =vy*t + \frac{1}{2}gt^2[/tex]

[tex]5 =12.8*t + \frac{1}{2}*9.8*t^2[/tex]

now by solving above quadratic equation we have

[tex]t = 0.345 s[/tex]

now in the same time the distance traveled horizontally is given as

[tex]d = v_x * t[/tex]

[tex]d = 15.3* 0.345 = 5.3 m[/tex]

so it will cover 5.3 m horizontally while it is dropped down by distance 5 m and it will take 0.345 s for the whole.