From a height of 25 meters a ball is thrown vertically upwards at a velocity of 5 meters per second.
a. What time in seconds will the ball reach its maximum height?
b. What is its maximum height?
c. When will the ball strike the ground?
d. What will be the ball’s velocity when it strikes the ground?
can someone help me out with this?
a.
Vi = 5
v = Vi - 9.8 t
v = zero at top
t = 5/9.81 seconds to top
b.
h = Hi + Vi t -4.9 t^2
= 25+ 5 t - 4.9 t^2
c.
0 = 25 + 5 t - 4.9 t^2
solve quadratic for t
use the positive t
the negative t is before you threw it :)
d.
again v = Vi - 9.81 t
v = 5 - 9.81*t
use t from part c
v will be negative (down) of course
oh, since this is calculus and not physics:
d^2h/dt^2 = -9.81
so
dh/dt = -9.81 t + constant
at t = 0, v = Vi
so
dh/dt = Vi -9.81 t
then
h = Vi t -(1/2)9.81 t^2 + constant
h = Hi when t = 0
so
h = Hi + Vi t -4.9 t^2
Sure! I can help you with that. To solve these questions, we can use the equations of motion for free-falling objects and a few basic principles of physics. Let's break down each question and explain how to find the answers step by step:
a. To find the time when the ball reaches its maximum height, we need to use the equation of motion for vertical displacement. The equation is:
h = ut + (1/2)gt^2
where:
h is the vertical displacement or height,
u is the initial velocity (in this case, the upward velocity of 5 m/s),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time.
In this case, we want to find the time at the maximum height, so the vertical displacement is 25 meters (the initial height). We know the initial velocity (u) is 5 m/s, and the acceleration due to gravity (g) is -9.8 m/s^2 because the ball is moving upward.
Plugging in the values, we have:
25 = (5)t + (1/2)(-9.8)t^2
This equation is a quadratic equation. To solve it, we can rearrange it and solve for t using either the quadratic formula or factoring method. Once you find the values of t, you can select the positive value since time cannot be negative.
b. To find the maximum height, we need to use the equation of motion for final velocity. The equation is:
v = u + gt
At the maximum height, the final velocity is zero because the ball momentarily stops before falling back down. We know the initial velocity (u) is 5 m/s, and the acceleration due to gravity (g) is -9.8 m/s^2.
Plugging in these values, we have:
0 = 5 + (-9.8)t
Solving for t will give us the time when the ball reaches its maximum height. Substituting this value of t into the equation of motion for vertical displacement:
h = ut + (1/2)gt^2
will give us the maximum height (h).
c. To find when the ball will strike the ground, we need to find the total time of flight. The total time of flight is the time it takes for the ball to go up, reach its maximum height, and then fall back down to the ground. The equation for the total time of flight is:
T = 2t
where t is the time we found in part a.
d. Finally, to find the ball's velocity when it strikes the ground, we can use the equation of motion for final velocity. The equation is the same as in part b:
v = u + gt
At the moment the ball strikes the ground, the final velocity will be the magnitude of the initial velocity because it falls in the same direction. So, we can use the equation with the initial velocity (u) and acceleration due to gravity (g) to find the velocity (v) at the time it strikes the ground.
I hope this breakdown helps you solve the problem! Feel free to ask if you have any further questions.