C the correct but not sure?
Solving this using the time, we know that range = horizontal velocity x time of flight
since
there are no horizontal forces acting on the ball, there are no
horizontal accelerations and the initial horizontal velocity of 36 cos
28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have
range = 36 cos 28 x 3.44 s = 109.3 m
Answer:
put these numbers in the boxes from up to down. hope this helps! :)
Explanation:
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Answer:
Explanation:
Given
Initial velocity u = 200m/s
Final velocity = 4m/s
Distance S = 4000m
Required
Acceleration
Substitute the given parameters into the formula
v² = u²+2as
4² = 200²+2a(4000)
16 = 40000+8000a
8000a = 16-40000
8000a = -39,984
a = - 39,984/8000
a = -4.998m/s²
Hence the acceleration is -4.998m/s²
When air resistance<span> acts, acceleration during a fall </span>will<span> be less than g because </span>air resistance affects<span> the motion of the falling </span>objects<span> by slowing it down. </span>Air resistance<span> depends on two important factors - the</span>speed<span> of the </span>object<span> and its surface area. Increasing the surface area of an </span>object<span> decreases its </span>speed<span>.</span>
