Question

John kicks the ball and ball does projectile motion with an angle of 53° to gorizontal its initial velocity is 10 m/s. Find the maximum height it can reach, horizontal displacement and total required for his motion

Answer 1

Answer:

Horizontal range is 9.6 m, Maximum height is 3.2 m and total time of motion is 1.6 s

Explanation:

As we know that John kicks the ball at speed v = 10 m/s

so we have

$v_x = 10 cos53 = 6 m/s$

$v_y = 10 sin53 = 8 m/s$

now we know that total time of the motion is given as

$T = \frac{2v_y}{g}$

$T = \frac{2(8)}{10}$

$T = 1.6 s$

Maximum height of the ball is given as

$Y_{max} = \frac{v_y^2}{2g}$

$Y_{max} = \frac{8^2}{20}$

$Y_{max} = 3.2 m$

Maximum range of the horizontal displacement is given as

$R = v_x T$

$R = (6)(1.6)$

$R = 9.6 m$

#Learn

Topic : Projectile Motion

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