The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.
The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.
v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s
Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.
The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.
Combining this together we get:
(1) vx=40m/s and vy=10m/s
Answer:
30.63 m
Explanation:
From the question given above, the following data were obtained:
Total time (T) spent by the ball in air = 5 s
Maximum height (h) =.?
Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:
Total time (T) spent by the ball in air = 5 s
Time (t) taken to reach the maximum height =.?
T = 2t
5 = 2t
Divide both side by 2
t = 5/2
t = 2.5 s
Thus, the time (t) taken to reach the maximum height is 2.5 s
Finally, we shall determine the maximum height reached by the ball as follow:
Time (t) taken to reach the maximum height = 2.5 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
h = ½gt²
h = ½ × 9.8 × 2.5²
h = 4.9 × 6.25
h = 30.625 ≈ 30.63 m
Therefore, the maximum height reached by the cannon ball is 30.63 m
I think it 32, but i’m not sure
The mass of an object always stays the same since it is really just the amount of matter in an object so no matter the force applied, as long as the object does not lose or gain matter, the object stays the same
