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kaheart [24]
3 years ago
12

A soccer ball is kicked horizontally off a bridge with a heigh of 36m. The ball travels 25m horizontally before it hits the pave

ment below. What was the soccer ball's speed when it was first kicked?
Physics
1 answer:
kow [346]3 years ago
5 0

Answer:

The initial velocity was U=22.14m/s

Explanation:

Step one :

Applying the third equation of motion

v² = u²+ 2as

Where v= Final velocity

U =initial velocity

a= acceleration due to gravity

S= distance or displacement

Step two :

V= 0

a= 9.81m/s²

S=25m

U=?

Step three :

Substituting into the equation we have

0²=U²+2*9.81*25

0=U²+490.5

U²=-490.5

U=√490.5

U=22.14m/s

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Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it
stich3 [128]

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

v^{2}=u^{2}-2a(y-2.2)\\

if we substitute values we have

v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m

c. the time it takes to arrive at 1.83m is obtain by using the equation below

1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37

if we insert the values, we solve for t , hence t=1.43secs

6 0
3 years ago
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