We can first obtain time of flight from vertical fall
Initial velocity U=0, d = 6 m, a = 9.8 m/s²
d = ut + 1/2 at²
6.0 = 0 + (1/2 × 9.80 t²)
t = √(12/9.8)
= 1.106 sec
Horizontal velocity = Vh = Dh/t
= 24.0 /1.106 s
= 21.69 m/s
The ball was thrown at a speed of 21.69 m/s
Answer:
9.8 m/s^2
Explanation:
This is a constant that you should memorize
Answer:
12 cm
Explanation:
From what I got with my answer (8.192 cm) I went ahead and rounded it to the closest answer which is 12 cm. Hopefully im correct but if not, I apologize in advance.
No not all, because not all electrical field attract and repel
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed

, and an accelerated motion on the y-axis, with initial speed

and acceleration

:


where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring

Therefore:

which has two solutions:

is the time of the beginning of the motion,

is the time at which the projectile hits the ground.
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point: