A ball rolls horizontally off a table of height 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the ground?

Answer 1

As the ball rolls horizontally of a ta!ble (of height 1.4m) with a speed of #4"m/s"# ,the horizontal component of velocity will be #v_"hor"=(4cos0)"m/s"=4"m/s"# and vertical component will be #v_"ver"=(4cos90)=0"m/s"#

If the time required to reach the be t sec then by the equation of motion under gravity we can write

#h=u_"ver"xxt+1/2xxgxxt^2#

Taking #"acceleration due to gravity "# #g=9.8"m/"s^2 and h=1.4m# we get

#=>1.4=0xxt+1/2xx9.8xxt^2#

#=>t=sqrt(2/7) s#

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Answer 2

Nathan Ashcroft

Using the kinematic equation $d = v_i t + \frac{1}{2} a t^2$ , where $d$ is the distance (1.4 m), $v_i$ is the initial velocity (0 m/s), $a$ is the acceleration due to gravity (-9.8 m/s²), and $t$ is the time, we can solve for $t$ :

$d = \frac{1}{2} a t^2$ $1.4 = \frac{1}{2} (-9.8) t^2$ $2.8 = -4.9 t^2$ $t^2 = \frac{2.8}{-4.9}$ $t^2 = -0.5714$ $t = \sqrt{-0.5714}$

Since time cannot be negative, we discard the negative solution:

$t = \sqrt{0.5714} \approx 0.756$

Therefore, it takes approximately 0.756 seconds for the ball to reach the ground.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.