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?

Question

Benjamin Asiedu · Accepted Answer

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#

Nathan Ashcroft · Answer

undefined 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.