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phy_2.8_projectile motion 拋體運動 2

Total questions: 10

Worksheet time: 50mins

Name
Class
Date
1.

一個物體斜向射往空中。物體到達最高點時,下列哪項敘述是正確的?

(1) 物體的垂直速度是零。

(2) 物體的水平速度是零。

(3) 物體的加速度是零。

An object is projected into the air at an angle. Which of the following statements is/are correct when the object is at its highest position?

(1) Its vertical velocity is zero.

(2) Its horizontal velocity is zero.

(3) Its acceleration is zero.

a)

只有 (1)

b)

只有 (2)

c)

只有 (1) 和 (2)

d)

(1)、(2) 和 (3)

2.

小球以投射角30° 從懸崖頂向上擲出,初速率為20 m s-1。球的初始水平速率 uxu_x 是多少?
A ball is projected at an angle of 30° above the horizontal from the top of a cliff with a speed of 20 m s-1. What is the initial horizontal speed uxu_x ?

a)

 20 ms120\ ms^{-1}  

b)

 20sin30° ms120\sin30\degree\ ms^{-1}  

c)

 20cos30° ms120\cos30\degree\ ms^{-1}  

d)

 20tan30° ms120\tan30\degree\ ms^{-1}  

3.

小球以投射角30° 從懸崖頂向上擲出,初速率為20 m s-1。5 s後,球的水平位移 sxs_x 是多少?
A ball is projected at an angle of 30° above the horizontal from the top of a cliff with a speed of 20 m s-1. What will be its horizontal displacement sxs_x  5 s later?

a)

 20×5 m20\times5\ m  

b)

 20sin30°×5 m20\sin30\degree\times5\ m  

c)

 20cos30°×5 m20\cos30\degree\times5\ m  

d)

 20tan30°×5 m20\tan30\degree\times5\ m  

4.

小球以投射角30° 從懸崖頂向上擲出,初速率為20 m s-1。球的初始垂直速率 uyu_y  是多少?
A ball is projected at an angle of 30° above the horizontal from the top of a cliff with a speed of 20 m s-1. What is the initial vertical speed uyu_y  ?

a)

 20 ms120\ ms^{-1}  

b)

 20sin30° ms120\sin30\degree\ ms^{-1}  

c)

 20cos30° ms120\cos30\degree\ ms^{-1}  

d)

 20tan30° ms120\tan30\degree\ ms^{-1}  

5.

小球以投射角30° 從懸崖頂向上擲出,初速率為20 m s-1。5秒後小球的垂直位移 sys_y  是多少?
A ball is projected at an angle of 30° above the horizontal from the top of a cliff with a speed of 20 m s-1. What will be its vertical displacement sys_y  5 s later?

a)

 20×5+12×9.81×52 m20\times5+\frac{1}{2}\times9.81\times5^2\ m  

b)

 20sin30°×5+12×9.81×52 m20\sin30\degree\times5+\frac{1}{2}\times9.81\times5^2\ m  

c)

 20sin30°×512×9.81×52 m20\sin30\degree\times5-\frac{1}{2}\times9.81\times5^2\ m  

d)

 20sin30°×5 m20\sin30\degree\times5\ m  

6.

小球從地面以投射角75° 擲出,擊中地面上的一個目標。若在同一位置以相同速率擲出小球,以下列哪個投射角擲出,可擊中同一個目標?

A ball is projected at an angle of 75° from the ground. It hits a target at the ground level. If the ball is projected from the same position with the same speed again, which of the following angle of projection would also enable the ball to hit the target?

a)

b)

12.5°

c)

15°

d)

45°

7.

小球從地面以速率10 m s–1和仰角30° 投出。當它到達度最大高度時,以下哪個物理量等於零?
A ball is fired from the ground with a speed of 10 m s-1 at an angle of 30° above the horizontal. Which of the following physical quantities is equal to zero when it is at the highest position. 

a)

動能 KE

b)

加速度 acceleration a

c)

速率 speed v

d)

垂直速率 vertical speed vyv_y  

8.

小球從地面以速率10 m s–1和仰角30° 投出。當它到達最大高度時,動能是多少?
A ball is fired from the ground with a speed of 10 m s-1 at an angle of 30° above the horizontal. What is its KE when it is at the highest position. 

a)

 12m(102) J\frac{1}{2}m\left(10^2\right)\ J  

b)

 12m(10sin30°)2 J\frac{1}{2}m\left(10\sin30\degree\right)^2\ J  

c)

 12m(10cos30°)2 J\frac{1}{2}m\left(10\cos30\degree\right)^2\ J  

d)

 00  

9.

小球從地面以速率10 m s–1和仰角30° 投出。當它到達最大高度時,以下哪條公式可用來計算其飛行時間 tHt_H ?
A ball is fired from the ground with a speed of 10 m s-1 at an angle of 30° above the horizontal. When it is at the highest position, which of the following equations can be used to find its time of flight  tHt_H ?

a)

 0=109.81×tH0=10-9.81\times t_H  

b)

 0=10sin30°9.81×tH0=10\sin30\degree-9.81\times t_H  

c)

 0=10sin30°+9.81×tH0=10\sin30\degree+9.81\times t_H  

d)

 10sin30°=10sin30°9.81×tH-10\sin30\degree=10\sin30\degree-9.81\times t_H  

10.

小球從地面以速率10 m s–1和仰角30° 投出。以下哪條公式可用來計算最大高度H?
A ball is fired from the ground with a speed of 10 m s-1 at an angle of 30° above the horizontal. Which of the following equations can be used to find its maximum height H?

a)

 12m(102)=mgH\frac{1}{2}m\left(10^2\right)=mgH  

b)

 12m(102102sin230°)=mgH\frac{1}{2}m\left(10^2-10^2\sin^230\degree^{ }\right)=mgH  

c)

 12m(102cos230°102)=mgH\frac{1}{2}m\left(10^2\cos^230\degree-10^2\right)=mgH  

d)

 12m(102102cos230°)=mgH\frac{1}{2}m\left(10^2-10^2\cos^230\degree\right)=mgH