y = y0 + v0*t - 1/2*g*t^2
where y0 is the initial height = 0, v0 is the initial speed, and g is the acceleration due to gravity, t is time, and y is the vertical position.
The ball reaches it's maximum height when dy/dt = 0 = v0 - g*t
and y = 5, so
5 = v0*t - 1/2*g*t^2
Use algebra to find the initial velocity and time.
The final velocity of the ball is given by
1. v = v0 - g*t, occurs when y = 0, or when
2. 0 = v0*t - 1/2*g*t^2
solve 2nd equation for t, and plug into 1st equation to solve for the final speed v at this time.
a ball is thrown up and reaches a maximum height of 5 m.
a) what is the initial velocity of the ball?
b) how long does it take for the ball to reach its maximum height?
c) what is the final velocity of the ball? if i returns to where it was thrown from?
2 answers
a sandbag is dropped from an unknown height. h, from the ground. if after 5 sec. it is located one half the height from which is thrown , at what height was it dropped?
b. what is its speed after 5 sec.?
c. what is its speed when it strikes the ground?
b. what is its speed after 5 sec.?
c. what is its speed when it strikes the ground?