U= 50m/s
g= -10m/s^2

The velocity at the highest point will be 0 m/s
Thus v=0

Height can be calculated by

2gs=v^2-u^2
-20s= -2500
s= 125m

Time :
v= u+gt
0=50-10t
-50=-10t
t= 5sec

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Initial Conditions:
The particle is thrown up vertically with a velocity of 50 m/s. We need to determine the velocity at the highest point, the height it reaches, and the time it takes to reach the highest point.

Velocity at the Highest Point:
At the highest point, the particle momentarily comes to rest before starting to fall downwards. Therefore, the velocity at the highest point will be 0 m/s.

Height Reached:
To find the height the particle reaches, we can use the kinematic equation:

v² = u² + 2as

where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (50 m/s)
a = acceleration (due to gravity, which is approximately -9.8 m/s²)
s = displacement (height reached)

Rearranging the equation to solve for s, we have:

s = (v² - u²) / (2a)

Substituting the values, we get:

s = (0² - 50²) / (2 * -9.8)
s = (-2500) / (-19.6)
s ≈ 127.55 m

Therefore, the particle will rise approximately 127.55 meters.

Time to Reach the Highest Point:
To find the time it takes for the particle to reach the highest point, we can use the kinematic equation:

v = u + at

where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (50 m/s)
a = acceleration (due to gravity, which is approximately -9.8 m/s²)
t = time

Rearranging the equation to solve for t, we have:

t = (v - u) / a

Substituting the values, we get:

t = (0 - 50) / (-9.8)
t = (-50) / (-9.8)
t ≈ 5.10 seconds

Therefore, it will take approximately 5.10 seconds for the particle to reach the highest point.

Summary:
- The velocity at the highest point is 0 m/s.
- The particle will rise approximately 127.55 meters.
- It will take approximately 5.10 seconds for the particle to reach the highest point.