<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities
sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s
cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s
Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s.
To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So
0.50 s * 14.7447368 m/s = 7.372368399 m
1.00 s * 14.7447368 m/s = 14.7447368 m
1.50 s * 14.7447368 m/s = 22.1171052 m
2.00 s * 14.7447368 m/s = 29.48947359 m
Rounding the results to 1 decimal place gives
0.50 s = 7.4 m
1.00 s = 14.7 m
1.50 s = 22.1 m
2.00 s = 29.5 m</span>
Answer:
550 kg
Explanation:
mass = E / gh
= 33000/60
=550
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consider the motion of projectile A in vertical direction :
v₀ = initial velocity of projectile A in vertical direction = 0 m/s (since the projectile was launched horizontally)
a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²
t = time of travel for projectile A = 3.0 seconds
Y = vertical displacement of projectile A = height of the cliff = h = ?
using the kinematics equation along the vertical direction as
Y = v₀ t + (0.5) a t²
h = (0) (3.0) + (0.5) (9.8) (3.0)²
h = 44.1 m
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
