# Projectile AP Physics 1?

A target lies flat on the ground 6 m from the side of a building that is 10 m tall, as shown below.The acceleration of gravity is 10 m/s^2. Air resistance is negligible.

A student rolls a 6 kg ball off the horizontal roof of the building in the direction of the target. The horizontal speed v with which the ball must leave the roof if it is to strike the target is most nearly (not 1 or 9)

### 5 Answers

- SpacemanLv 71 month agoFavorite Answer
g = acceleration of gravity = 10 m/s²

y = height of building = 10 m

x = target distance from building = 6 m

v0y = initial vertical velocity = 0 m/s

vy = final vertical velocity = to be determined

v0x = initial horizontal velocity = to be determined

(vy)² = (v0y)² + 2gy

(vy)² = (0 m/s)² + 2(10 m/s²)(10 m)

(vy)² = 200 m²/s²

vy = 14.14213562 m/s Final Vertical Velocity Component

vy = v0y + gt

vy - v0y = gt

(vy - v0y) / g = t

t = (vy - v0y) / g

t = (14.14214 m/s - 0 m/s) / 10 m/s²

t = 1.414213562 s Time for Ball to Fall 10 meters

v0x = x / t

v0x = 6 m / 1.627697 s

v0x = 4.242640687 m/s NECESSARY INITIAL HORIZONTAL VELOCITY

v0x = 3√2 Answer (2)

- oubaasLv 71 month ago
falling time t = √2h/g = √2 sec

speed V = d/t = 6/√ 2 = 6√ 2 /2 = 3√ 2 m/sec ..answer 2

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