Answer:
15.7 m/s
Explanation:
The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:

where
u = 0 is the initial velocity
g = 9.8 m/s^2 is the acceleration
t is the time
If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:

Answer:
- 3 meter
Explanation:
A dog has started motion from +3 meter. ...(Given)
∴ maximum positive distance = + 3 meter
Magnitude of distance = 3
Maximum negative distance = (-) (magnitude of distance)
Maximum negative distance = (-) (3)
Maximum negative distance = -3 meter
Hence, maximum negative distance is -3 meter.
Answer:
In February 1784, just after the close of the Revolutionary War, the General Assembly of Georgia earmarked 40,000 acres of land to endow "a college or seminary of learning." The following year, Abraham Baldwin, a lawyer and minister educated at Yale University in New Haven, Connecticut, who had settled in Georgia
Explanation:
please mark this answer as brainliest
kinetic energy is given as
KE = (0.5) m v²
given that : v = speed of the bottle in each case = 4 m/s
when m = 0.125 kg
KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J
when m = 0.250 kg
KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J
when m = 0.375 kg
KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J
when m = 0.0.500 kg
KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>