Physics question! Please Help?

h max = Vy^2/2g = 11.7^2/19.612 = 6.980 m

v = u - gt

Rock will reach maximum height when v = 0

- u = -gt

t = u/g

y = ut - 1/2gt^2

At maximum height, t = u/g

H = u(u/g) - 1/2g (u/g)^2

= u^2/g - 1/2(u^2/g)

= 1/2u^2/g

= 0.5x11.7^2)/9.8

= 6.984183672

= 6.98 m (2 dp)

Given,

Mass (m) = 0.78 kg

Final velocity (v) = 0m/sec [∵ at max height]

Initial velocity (u) = 11.7 m/sec

Maximum height (hmax) = ?

Using kinematic relation:-

v = u - gt [∵ direction of rock is upwards]

→ u = gt

→ t = u/g

Now, hmax = ut - ½gt² [∵ direction of rock is upwards]

hmax = u x [u/g] - ½ x 9.8 x (u/g)²

= (u²/g) - 4.9 [u/g]²

= u²/g - 4.9 x [u²/g²] = u²/g [1-4.9 x 1/g]

= (11.7)² / 9.8 [1-4.9 x 1/9.8] = 6.98 m

∴ maximum height (hmax) = 6.98 m.

As the rock rises to its maximum height, its velocity decreases from 11.7 m/s to 0 m/s. The acceleration is -9.8 m/s^2. Let’s use the following equation to determine the distance the rock rises.

vf^2 = vi^2 + 2 * a * d
0 = 11.7^2 + 2 * -9.8 *
d = 136.89 ÷ 19.6

The maximum height is approximately 6.98 meters. I hope this is helpful for you.