# 高中物理題目鉛直上拋?

1.從頂樓鉛直上拋一石，若經頂樓上方h公尺石速率為經過頂樓下方h公尺時的一半，則此物可爬升之最高點具頂樓為幾公尺

2.甲乙兩球分別以V,2V的初速，同時同地鉛直上拋，重力加速度g，當甲球升到最高點瞬間，甲乙兩球之間的高度差為多少

3將兩球同時同地鉛直上拋，若兩球上升最大高度差為35公尺，落地時間差2秒，重力加速度g=10m/s，則兩球拋出的初速度差為多少m/s

### 1 個解答

• 最佳解答

Q.1.

Take upwards be the positive direction

Let v m/s be the velocity of the stone at the height h m above the building, then -2v m’s is the velocity of the stone at the height h m below the building.

Consider the stone from the position of h m above the building to h m below the building

By using v^2 = u^2+2as

(-2v)^2 = v^2+2(-g)(-2h)

4v^2 =v^2+4gh

3v^2 = 4gh

v^2 = 4gh/3

Consider the maximum displacement be s m

Start from the stone at the height h m above the building to the maximum level

By using v^2 = u^2+2as

0^2 = v^2+2(-g)(s – h)

0 = v^2-2(g)(s – h)

s – h = v^2/2g

s = v^2/2g + h

s = 4gh/3(2g) + h (As v^2 = 4gh/3)

s = 5h/3 m

Q.2.

Take upwards be the positive direction

Let s1 m be the maximum height of ball A reached, then s2 m is the height of ball B when ball A reaches its maximum height

For Ball A

By using v^2 = u^2+2as

0^2 = v^2+2(-g)(s1)

0 = v^2-2gs1

s1 = v^2/2g m

Time for Ball A reaches its maximum height

By using s = 0.5(u + v)t

S1 = 0.5(v + 0)t

v^2/2g = 0.5vt

t = v/g sec

For the height of Ball B travels through this time period

By using s = ut + 0.5at^2

S2 = (2v)(v/g) + 0.5(-g)(v/g)^2

S2 = 3v^2/2g m

So, the difference in height

= 3v^2/2g - v^2/2g

=2v^2/2g

= v^2/g m

Q.3

We can use Q.2. Ball A as the reference

Maximum height s = v^2/2g m

And time to reach maximum height t = v/g sec

So time to return the ground t = 2v/g sec

Let s1 and s2 be the maximum height of ball 1 and ball 2 respectively

Let v1 and v2 be the initial velocities of ball 1 and ball 2 respectively,

While t1 and t2 be the times of ball 1 and ball 2 to return the ground respectively. Given that ball 2 is faster than ball 1.

S1 = v1^2/2g and S2 = v2^2/2g

So s2 – s1 = v2^2/2g - v1^2/2g = 35

v2^2 – v1^2 = 70g = 700 ………….. (1)

t1 = 2v1/g and t2 = 2v2/g

so t2 – t1 = 2v2/g - 2v1/g = 2

v2 – v1 = g = 10 ……………. (2)

From (1)

v2^2 – v1^2 = 70g = 700

(v2 – v1)( v2 + v1) = 700

v2 + v1 = 70

v2 + v2 + 10 = 70 (By (2))

v2 = 40 m/s , so v1 = 30 m/s

• 登入以對解答發表意見