Standing on the top ledge of a 55 m high building you throw a ball straight up with an initial speed of 29 m/s. How long to the nearest second, does it take to hit the ground?
Answer after rounding is 7 seconds.
The next question is "If the building in the previous problem is 44 meters high and you throw the ball up at 30 m/s, how high to the nearest meter does it go?"
I don't understand how to figure out what the highest meter is.
Apparently you have already answered question 1. If not, see the "related Question" below.
Question 2:
If the ball starts at a height of Ho = 44 m, if loses the available kinetic energy until it reaches the maximum height Hmax
g*Ho + Vo^2/2 = g*Hmax
(Hmax-Ho)= Vo^2/(2g) = 45.9 m
Hmax = 89.9 m
To find out how high the ball goes when thrown up, you can use the equations of motion. In this case, since the ball is thrown vertically upward, we can use the kinematic equation:
Final velocity squared = Initial velocity squared + 2 * acceleration * displacement
In this equation, the final velocity will be 0 m/s (because the ball comes to a stop at the highest point), the initial velocity is 30 m/s, and the acceleration is -9.8 m/s^2 (since gravity is acting in the opposite direction).
Now, let's substitute these values into the equation:
0^2 = (30^2) + 2 * (-9.8) * displacement
Simplifying this equation, we have:
0 = 900 - 19.6 * displacement
Rearranging the equation to solve for displacement, we get:
Displacement = 900 / 19.6
Calculating this, we find:
Displacement = 45.91 meters
Therefore, the ball reaches a height of approximately 45.91 meters when thrown up at 30 m/s. Rounding to the nearest meter, the answer is 46 meters.