Aball is thrown horizontally from the top of
a building 82 m high. the ball strikes the
ground 53 m horizontally from the point of
release.
what is the speed of the ball just before it
strikes the ground?
answer in units of m/s.

s = ut + at²/2

82 = 0*t = 10*t²/2

t = 4.0s

s=ut

53 = H*4

H=13.25 m/s   - Horizontal velocity

v² = u²+2as

v² = 0 + 2*10*82

v =40.5 m/s     - vertical velocity

vector sum

R²= V²+H²

R² = 40.5²+13.25²

R = 42.6m/s

Explanation:

horizontal velocity of the ball stays constant throughout the time because there is no horizontal forces acting on the ball to change its velocity ( according to newtons second law of motion). but the vertical velocity of the ball will increase (initially it is zero) because the gravitational force is acting vertically downward upon the ball. so the final velocity will be the vector sum of horizontal and vertical velocities at the point of impact with the ground. we already know the horizontal distance(53m) and vertical distance (82m) that the ball travels. there is no acceleration horizontally but the vertical acceleration is g = 9.8 m/s²(10m/s² approximately). so first from the vertical movement we can find the time of flight of the ball to hit the ground by the equation (s = ut + at²/2) . and also the vertical velocity at impact can also be found( v² = u²+2as) . to find the horizontal velocity we can use (s=ut) here t is the time of flight of the ball.

after finding the two velocities we can get the vector sum of the two. the velocities are normal to each other.

so the final velocity of the ball R will be

R² = V²+H²        V=vertical velocity, H = Horizontal velocity

the answer would be a

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