a) A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 100 m above ground level, and the ball is fired with initial horizontal speed v_0. Assume acceleration due to gravity to be g = 9.80 m/s^2. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time tg. What is the y position of the cannonball at the time tg/2?

I got 75m and I know that one is right.

I need help with part b
Given that the projectile lands at a distance = 110 from the cliff, as shown in the figure, find the initial speed of the projectile, .

If you found tg, the time in the air, then

110=horizontalvelocity*tg

I found tg to be 4.5. Is it 24.44

yes, watch the significant digits.

To find the initial speed of the projectile, we can use the horizontal motion equation. In this case, the only horizontal force acting on the projectile is the initial velocity, since there is no horizontal acceleration.

The horizontal distance traveled by the projectile can be calculated using the formula:
s_x = v_0 * t_g
where s_x is the horizontal distance, v_0 is the initial horizontal speed, and t_g is the time it takes for the projectile to hit the ground.

Given that the horizontal distance traveled is equal to 110 m, we can write the equation as:
110 = v_0 * t_g (equation 1)

Now, let's consider the vertical motion of the projectile. The vertical displacement can be calculated using the formula:
s_y = v_0 * t - 1/2 * g * t^2
where s_y is the vertical displacement, v_0 is the initial vertical speed, t is the time, and g is the acceleration due to gravity.

Since the projectile hits the ground at time t_g, we can write the equation as:
0 = v_0 * t_g - 1/2 * g * t_g^2 (equation 2)

We need to use equation 2 to solve for v_0. Rearranging the equation, we get:
v_0 = 1/2 * g * t_g (equation 3)

Now we can substitute equation 3 back into equation 1 to find the initial speed of the projectile:
110 = (1/2 * g * t_g) * t_g
110 = 1/2 * g * t_g^2
t_g^2 = 2 * 110 / g
t_g^2 = 220 / g

Since we have the value of g (9.8 m/s^2), we can find t_g:

t_g^2 = 220 / 9.8
t_g^2 = 22.45
t_g ≈ √22.45
t_g ≈ 4.74 seconds (approx.)

Now, substitute the value of t_g into equation 3 to find v_0:

v_0 = 1/2 * g * t_g
v_0 = 1/2 * 9.8 * 4.74
v_0 ≈ 23.15 m/s (approx.)

Therefore, the initial speed of the projectile is approximately 23.15 m/s.