A bullet is launched with an angle and initial velocity ν = 1.4 x 103 m/sec and reaches the target at a horizontal distance of 2 x 105 m. When the acceleration due to gravity is 9.8 m/s2 then the launch angle is n, where n is….
A. 10
B. 30
C. 45
D. 60
E. 75
→Peluru ditembakkan condong ke atas dengan kecepatan awal ν = 1,4 x 103 m/s dan mengenai sasaran yang jarak mendatarnya sejauh 2 x 105 m. Bila percepatan gravitasi 9,8 m/s2, maka elevasinya adalah n derajad dengan n sebesar…
A. 10
B. 30
C. 45
D. 60
E. 75
Sumber soal : UMPTN 1993

Discussion

The horizontal range (maximum distance) of bullet trajectory R = xmax = 2 x 105 m, g = 9.8 m/sec2, initial velocity νo = ν = 1,4 x 103 m/sec.

 

In Fig. 4-36, a stone is projected at a cliff of height h with an initial speed of 42.0 m/s directed at angle θo = 60.0° above the horizontal. The stone strikes at A, 5.50 s after launching. Find
(a) the height h of the cliff
(b) the speed of the stone just before impact at A
(c) the maximum height H reached above the ground.

Discussion

(a) the height h of the cliff
h = νo cos θ t
h = νo cos θ t
h = (42.0) (0.5)(5.5)
h = 115.5 m

(b) the speed of the stone just before impact at A
horizontal speed:
νtx = νo cos θ
νtx = (42.0)(0.5) = 21 m/s

vertical speed:
νty = νo sin θ − gt
νty = (42.0)(0.5√3)− (9.8)(5.5) = 35.7− 53.9 = − 18.2 m/s

the speed:
νt = √[(ty)2 +(tx)2]
proceed to get the result.

Toni throws a ball toward a wall at speed 20.0 m/s and at angle θo = 37.0° above the horizontal. The distance of wall is d = 24.0 m from the release point of the ball.
(a) How far above the release point does the ball hit the wall?
(b) What is the horizontal component of its velocity as it hits the wall?
(c) What is the vertical component of its velocity as it hits the wall?
(d) When the ball hits, has it passed the highest point on its trajectory?

Use sin 37° = 0.6 and acceleration due to gravity 10.0 m/s2
→ Toni melempar sebuah bola ke arah dinding di depannya dengan kelajuan 20 m/s dan sudut θo = 37.0° dari garis mendatar. Jarak dinding adalah 24 m dari titik lempar bola.
(a) Berapa ketinggian bola saat mengenai dinding dihitung dari titik lempar?
(b) Berapakah kecepatan horizontal bola saat mengenai dinding?
(c) Berapakah kecepatan vertikal bola saat mengenai dinding?
(d) Saat mengenai dinding apakah bola telah melampaui titik tertingginya?

Discussion

(a)How far above the release point does the ball hit the wall?
Find the time t when ball hits the wall from horizontal distance x:
x = vo cos 37° t
24.0 = (20)(0.8)t
t = 1.5 sec

Then we have the height y
y = vo sin 37° t − 1/2 gt2
y = (20)(0.6) (1.5) − 1/2 (10)(1.5)2
y = 6.75 m

(b) What is the horizontal component of its velocity as it hits the wall?
The horizontal component of ball velocity:
vtx = vox = vo cos 37°
vtx = (20)(0.8) = 16 m/s

(c) What is the vertical component of its velocity as it hits the wall?
The vertical component of ball velocity :
vty = vo sin 37° − gt
vty = (20)(0.6) − (10)(1.5) = −3 m/s

(d) When the ball hits, has it passed the highest point on its trajectory?
The time to reach highest point ymax is:
t = vo sin 37°/g = (20)(0.6)/10 = 1.2 sec, the time to reach te wall is 1.5 sec, so the conclusion is the ball has passed the highest point.

(c) the maximum height H
H = 2o sin2 θ )/2g
proceed to get your own result.

 

A car is going to travel across a ditch with 4.0 m of width. The height difference between the two sides of the ditch is 15 cm. Find the minimum speed of the car to reach the left side successfully!

→ Sebuah mobil hendak menyeberangi sebuah parit yang lebarnya 4,0 meter. Perbedaan tinggi antara kedua sisi parit itu adalah 15 cm seperti yang ditunjukkan oleh gambar di atas ini. Jika percepatan gravitasi g = 10 m/s2, maka besar kelajuan minimum yang diperlukan oleh mobil tersebut agar penyeberangan mobil itu dapat berlangsung adalah….
A. 10 m/s
B. 15 m/s
C. 17 m/s
D. 20 m/s
E. 23 m/s
(Soal UMPTN 1998 Fisika)

Discussion

Finding the time to reach the left side of the ditch, use the vertical axis:
y = vo sin θ t + 1/2 gt2
y = 0 + 1/2 gt2
0.15 = 1/2 (10)t2
t = √0.003 = 0.1√3 sec
To reach 4.0 m, the car velocity must be
x = vo cos θ t
x = vo cos 0° t
4.0 = vo (1)(0.1√3)
vo = 23 m/s

 

A bullet is launched with an  initial velocity ν = 100 m/sec at an incline and launch angle of 60° see the figure

When the acceleration due to gravity is 10.0 m/s2 find the value of d where the bullet impacts the incline!
→Peluru ditembakkan condong ke atas dengan kecepatan awal ν = 100 m/s dan sudut elevasi 60° pada suatu bidang miring seperti gambar di atas. Bila percepatan gravitasi 10 m/s2 tentukan jarak d saat peluru mengenai bidang miring!
(Fisika Study Center)

Discussion

We will just use the shortcut for this problem, you could search  the calculation steps in indonesian version.

Where for this question:
θ = 60°
α = 30°

Then insert the data