Question

A soccer ball is kicked horizontally off a 22.0 meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

Answer 1

A)

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Answer : 16.5m/s

Given:

x=35.0m

ax=0m/s/s

y=−22.0m

viy=0m/s

ay=−9.8m/s/s

Use y=viy.t+0.5×ay×t2 to solve for time → the time of flight is 2.12 seconds.

Now use x=vix×t+0.×ax×t2 to solve for vix.

Note that ax is 0m/s/s so the last term on the right side of the equation cancels.

By substituting 35.0m for x and 2.12s for t the vix can be found to be 16.5m/s

Answer 2

Answer:

The initial horizontal velocity of soccer ball is 16.58m/s

Explanation:

Given that,

Height of hill(h)=22m

Horizontal Distance(X)=35m

Using equations of motion for a horizontally projected body,we find the time taken by soccer ball to land

$h = u_{y}t + \frac{1}{2} gt {}^{2} \\ h = 0 + \frac{1}{2} gt {}^{2} \\ = > 22 = \frac{1}{2} \times 9.81 \times t {}^{2} \\ = > t = 2.11sec$

Similarly,using equation for distance in horizontal direction,we find the initial horizontal velocity of soccer ball

$X = u_{x}t + \frac{1}{2} a_{x}t {}^{2} \\ = > X =u_{x} t \\ = > 35 =u_{x} \times 2.11 \\ = > u_{x} = 16.58ms {}^{ - 1}$

Therefore,the initial horizontal velocity of soccer ball is 16.58m/s

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