Abby asked in Science & MathematicsPhysics · 1 month ago

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)

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  • 1 month ago
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    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)

  • oubaas
    Lv 7
    1 month ago

    falling time t = √2h/g =  √2 sec 

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

  • 1 month ago

    V*t = 6m and 10= ½*10*t² so t = √2 and V = 6/√2 = 3√2 choice 2 <<<<

  • 1 month ago

    (As shown below picture) (wouldnt let me put more than 1 picture sorrry)

    Attachment image
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  • 1 month ago

    as shown below ??

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