Two projectiles are launched at 100m/s the angle of elevation for the first is 30 degrees the second is?

Question

60 degrees. Which projectile has the lower speed at max altitude? what is the ranges of projectiles? what is the acceleration of projectiles while in flight?

Answer 1

a) Lower speed ?
Given u1 = u2 = 100 m/s
Angles A1 = 30 and A2 = 60 degrees with the horizontal
At max altitude the upward component of velocity is zero in both cases. That is, v1y = v2y = 0.

Acceleration being acceleration of gravity g, acting vertically downwards, horizontal acceleration is zero.
v1x = u1x = u1 cos A1 = 100xcos 30 = 100x0.866 = 86.6 m/s.
v2x =u2x = u2 cos A2 = 100xcos 60 = 100x0.5 = 50.0 m/s.
Hence the second projectile launched at 60 degree has the lower speed at the max altitude.

b) Range R = ux*t
To determine time t, take Sy = 0, a = g = 9.8 m/s2, uy = u sin A
Using the equation, S= ut + 1/2 a t2
and Sy = 0; uy = u sin A and ay = g = 9.8 m/s2; t = 2uy/g
t1 = 2x u sin A /g = 2x100x0.5/9.8 = 100/9.8 =10.2 s
t2 =2x u sin A /g = 2x100x0.866/9.8 =2x 86.6/9.8 = 17.67s.

R1 = u1x* t1 = 100 cos 30 x t1 = 100x0.866x10.2 = 883.3 m.

R2 = u2x* t2 = 100 cos 60 x t2 = 100x0.5x17.67 = 883.5 m.

Alternately using the formula for range, R = (u.u/g) sin 2A
Sin 2A = Sin 60 and sin 120 = sin 60.
Thus, R1 = R2 = 100x100x0.866/9.8 = 884 m.

Note : The difference in the 3rd digit is due to rounding off and is permissible.

c) Neglecting air resistance, the only force acting on a projectile is force of gravity and hence the aceleration is acceleration of gravity = g = 9.8 m/s2 downward.

Answer 2

At maximum altitude the vertical velocity of both is zero, and the horizontal component is the actual velocity. It's lower for the 60 degree projectile. Acceleration in flight is g; 9.81 m/s^2 downwards. Horizontal velocity is constant, neglecting air resistance. To calculate range you work out how long each projectile is in the air, using v sin angle of elevation for the vertical component of velocity, and equation of motion. During this time it's moving across the ground at velocity v cos angle of elevation, so you can work out how far it's travelled by the time it hits the ground. You'll find that for a given launch speed you get max. range when angle of elevation=45 degrees.

Answer 3

Projectile launched at 60 degrees has less speed at max altitude

R = u^2 sin(2*angle)/g

Acceleration = g downwards.

Answer 4

i'm so sorry approximately your kitty! Kitties are very innovative approximately those circumstances. i'm constructive your cat would be superb. There are some probable opportunities: a million-it is going to proceed to be energetic at nighttime, producing adequate physique warmth to stay comfortable, 2-it gets chilly and are available homestead, or 3-it is going to detect a niche in the community it is heat...ex: somebody's storage... If I have been you, i might take some steps. bypass away kitty's prominent handle close to the place it exited your place. positioned some blankets there in a cardboard field. it is available to rig up somewhat outdoors heater employing an electric powered mild (verify it is hearth risk-free). i understand it will be annoying till your infant comes homestead... draw close in there.

Answer 5

Projectile A (30 degrees):
Vv=100*sine 30=50 m/s @ 0 altitude
Vh=100*cos 30=86.6 m/s @ max altitude
Flight time, t1=Vv/g
t1=50/10=5 seconds
Return time, t2=t1=5 seconds
Range, R=vh*(t1+t2)
R=86.6*10=866 m
vertical acc., av=g=-10 m/s^2
horizontal acc., ha= -vh/(t1+t2)
ha=-86.6/10= - 8.66 m/s^2
total acc.,a=sqrt((-8.66^2)+10^2)
a=13.2 m/s^2 @ 30 degrees
Projectile B (60 degrees):
Vv=100*sine 60=86.6 m/s @ 0 altitude
Vh=100*cos 30=50 m/s @ max altitude
Flight time, t1=Vv/g
t1=86.6/10=8.66 seconds
Return time, t2=t1=8.66 seconds
Range, R=vh*(t1+t2)
R=50*17.32=866 m
vertical acc., av=g=-10 m/s^2
horizontal acc., ha= -vh/(t1+t2)
ha=-50/17.32= - 2.9 m/s^2
total acc.,a=sqrt((-2.9^2)+10^2)
a=10.4 m/s^2 @ 60 degrees
Therefore, (i) projectile B has the lower speed at max. altitude (ii)Range of A=range of B= 866 m (iii) acceleration of A= -13.2 m/s^2 at 30 degrees and acceleration of B= -10.4 m/s^2 at 60 degrees.

Answer 6

Use components, its easy to solve it. At max altitude, the vertical component is 0. 100cos60 for the first, 100cos30 for the second.
the range is found by the formula (initial velocity to the square x sin(2x angle) divide by gravity(9.81 or10)

max range= (initial velocity to the square)divide by g

acc. :

-g x vertical component.

Answer 7

theoretically both should have the same speed.