Wiki User
ā 12y agoH = H0 + V0 T + 1/2 G T2
H0 = 70 m
V0 = 12 m/s
G = -9.8 m/s2
We want to know ' T ' when H=0 .
0 = 70 + 12 T - 4.9 T2
Quadratic formula:
T = -1/9.8 [ -12 ± sqrt(144 + 1,372) ] =
-1/9.8 [ -12 ± 38.9358 ] =
T = 5.1975 seconds(T = -2.7486 seconds is also a solution to the math but isn't physically relevant)
Wiki User
ā 12y agov2 = u2 + 2as where v = current velocity, u = initial velocity, a = acceleration, and s = displacement. Taking a = - 9.8 ms-2 v2 = 182 - (9.8 x 11 x 2) = 108.4 v = 10.4 ms-1
1000000 m
If that's 32.1 meters per second initially, then after 4 seconds it's fallingwith a speed of 7.1 meters per second.If that's 32.1 feet per second initially, then it returns to the thrower's hand injust under 2 seconds, and it's in the dirt long before 4 seconds have passed.If it had been tossed at the edge of a cliff, then after 4 seconds, it would befalling with a speed of 96.7 feet per second.
if the bal is thrown by making 45 degree angles. with the ground..it will travel maximum distance...
The ball was thrown horizontally at 10 meters per sec, and the thrower's arm was 78.4 meters above the base of the cliff.
The answer depends on whether the ball is thrown vertically upwards or downwards. That critical piece of information is not provided!
The speed of a ball thrown horizontally remains constant, so it will still be 9 meters per second one second later. Since there is no vertical acceleration, only the horizontal motion occurs.
10 meters per second due to gravity
The acceleration of an object thrown vertically upwards can be calculated using the kinematic equation (v_f^2 = v_i^2 + 2a \cdot d), where (v_f) is the final velocity, (v_i) is the initial velocity, (a) is the acceleration, and (d) is the distance. Given that the object is thrown vertically upwards, the equation becomes (0 = (44 , \text{m/s})^2 + 2 \cdot a \cdot (-3.5 , \text{m})). Solving for (a), we find that the acceleration is approximately -104 m/sĀ², which indicates that the object is accelerating downwards.
After a second, the ball will still have a horizontal velocity of 8 meters per second. It will also have a vertical velocity of 9.8 meters per second (Earth's acceleration is about 9.8 meters per square second). The combined speed (using the Law of Pythagoras) is about 12.65 meters per second.
A projectile fired directly upwards has no positive velocity. Its only velocity is attributed to the force of gravity, which is -9.8 meters per second squared.
At the highest point, the velocity of an object thrown vertically into the air is momentarily zero as it changes direction. This is the point where it transitions from going upward to downward.
It depends on the height of the building and also on the direction the object is thrown in (up, down etc.).
Horizontally
90
v2 = u2 + 2as where v = current velocity, u = initial velocity, a = acceleration, and s = displacement. Taking a = - 9.8 ms-2 v2 = 182 - (9.8 x 11 x 2) = 108.4 v = 10.4 ms-1
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?