How High Does a Ball Go When Thrown Upward at 25m/s?

[JBemp]

Homework Statement

A ball is thrown stright up with the speed 25m/s
how high does the ball go?

Homework Equations

S=v*t or V=vot+at
I know gravity is a neg acceloration so I should have -9.81 somewhere

The Attempt at a Solution

I need to find the time first am guessing so, T=s/v but i don't know S ether just 25m/s and -9.81 I know this is most likly very simple but it just eludes me

 

[Doc Al]

You need to make use of the kinematic equations for accelerated motion: Basic Equations of 1-D Kinematics

 

[m2003]

Based on the given information, the ball is thrown with an initial velocity of 25m/s and experiences a constant acceleration due to gravity of -9.81m/s^2. To find the height the ball reaches, we can use the equation h = vo*t + (1/2)at^2, where h is the height, vo is the initial velocity, t is the time, and a is the acceleration. Since the ball is thrown straight up, we can take the initial velocity to be positive 25m/s and the acceleration to be negative -9.81m/s^2.

To find the time, we can use the equation t = (vf - vo)/a, where vf is the final velocity (which is 0 since the ball reaches its highest point and then falls back down) and vo is the initial velocity. Plugging in the values, we get t = (0 - 25)/(-9.81) = 2.55 seconds.

Now, using the time we calculated, we can find the height by plugging in the values in the first equation. h = (25)(2.55) + (1/2)(-9.81)(2.55)^2 = 31.9 meters. Therefore, the ball reaches a height of approximately 31.9 meters before falling back down.