A soccer ball leaves a cliff 20.2 m above the valley ﬂoor, at an angle of 10 degrees above the horizontal. The ball hits the valley floor 3.0 seconds later. What is the initial velocity of the ball?

Question

Julia Aguilar · Accepted Answer

#46.05"m/s"#

The situation looks like this:

I will adopt a convention that "up is positive". This means that #y# will be -ve in this scenario.

We can use:

#s=ut+1/2at^2#

This becomes:

#y=v(sintheta)t-1/2"g"t^2#

#:.-20.2=3vsintheta-1/2xx9.8xx3^2#

#:.3vsintheta=44.1-20.2=23.9#

#:.v=(23.9)/(3sin10)#

#v=(23.9)/(3xx0.173)=46.05"m/s"#

Charlotte Aaron · Answer

undefined To find the initial velocity $ v_0 $ of the soccer ball, we can use the kinematic equation for horizontal motion:

$$ x = v_0 \cdot t \cdot \cos(\theta) $$

Given that $ x = 20.2 $ m, $ t = 3.0 $ s, and $ \theta = 10^\circ $, we can solve for $ v_0 $:

$$ v_0 = \frac{x}{t \cdot \cos(\theta)} $$

$$ v_0 = \frac{20.2 \, \text{m}}{3.0 \, \text{s} \cdot \cos(10^\circ)} $$

$$ v_0 \approx \frac{20.2 \, \text{m}}{3.0 \, \text{s} \cdot 0.9848} $$

$$ v_0 \approx \frac{20.2 \, \text{m}}{2.9544} $$

$$ v_0 \approx 6.84 \, \text{m/s} $$

So, the initial velocity of the soccer ball is approximately $ 6.84 \, \text{m/s} $.