Physics Question! A projectile is launched with an initial speed of 40 m/s at an angle of 35° above the horizo?
A projectile is launched with an initial speed of 40 m/s at an angle of 35° above the horizontal on flat ground. Neglect air friction. (a) What is the projectile's velocity at the highest point of its trajectory? m/s (b) Now suppose the projectile instead lands on a hillside 4.0 s after launch. What is the straight-line distance from where the projectile was launched to where it hits? (Note that the hill might slope up or down from the launch point.) m
A projectile is launched at some angle...?
A projectile is launched at some angle to the horizontal with some initial speed and air resistance is negligible. (a) What is its acceleration in the vertical direction? (Let up be the positive direction.) (b) What is its acceleration in the horizontal direction? Units for both should be in m/s^2 I know that the projectile is a freely falling body, but I have no clue how to solve this question. Thank you!
Projectile Motion Question?
the range equation is R = v0^2 sin(2*theta)/g the max height achieved is h = v0^2 sin^2(theta)/2g equating: v0^2 sin(2*theta)/g = v0^2 sin^2(theta)/2g cancel common terms of v0^2/g and get sin(2 theta) = sin^2 (theta)/2 use the trig identity sin(2theta) = 2 sin(theta) cos(theta) and we have 2 sin(theta) cos(theta) = sin^2(theta)/2 2 cos(theta) = sin(theta)/2 tan(theta) = 4 or theta = 76 deg
A projectile is launched upward... PHYSICS question!?
yo = 84 m vy = 56.5 m/s Time to vy=0 t = vy / g =5.765 s maximum height above the point of launch: y = yo + vy * t + (1/2) * g * t ^ 2 = 246.87 m h = y-yo = 162.87 m Flight time: td = Sqr(2 * y / g)= 7.098 s T = t + td = 12.863 s velocity at impact: vyf = g * td = -69.56 m/s
A projectile is launched at ground level with an initial speed of 52 m/s at an angle of 31º above the horizon?
1A) x=v0x*t=v0cosθ*t x=52co31*3.2=142.6 m 1B) y0=1/2gt^2-v0y*t=1/2gt^2-v0sinθt y=0.5*9.8*3.2^2-52*sin31*3,2=23.4 m 2A) x=2v0^2sin(2θ)/g v0=[xg/2sin(2θ)]^1/2=14.4 m/s the initial speed relative to the ground is v=v0-4.4=10 m/s 2B) fly time is t=2voy/g t=2*14.4/9.8=2.94 2C) mgy=1/2mv0y^2 y=v0y^2/(2g)=10.58 m
A projectile is launched upward at an angle of 75 degrees from the horizontal and strikes the ground a certain distance downrange. For what other angle of launch at the same speed would this projectile land just as far away?
Please try to solve your home work to learn butter and understand the concept wellIt is a simple ex exercise,you can solve be self dependent refer to any book of classical mechanics,projectile chapter.
A projectile is launched upward at an angle of 75° from the horizontal and strikes the ground a certain distance down range. At what other angle of launch at the same speed would this projectile land just as far away?
A projectile launched at an angle [math]\theta[/math] with respect to the horizontal will have the same range as a projectile launched at an angle [math]\frac{\pi}{2} - \theta[/math].For [math]\theta = \frac{5\pi}{12} \qquad \theta_c = \frac{\pi}{12}[/math] or 15 degrees.
A projectile is fired with speed 40m/s at an angle of 30° with the horizontal. What is the range of projectile?
Speed of the projectile = 40 m/sAngle with the horizontal at which it is fired = 30°Component of the projectile velocity along the horizontal direction= 40 m/s × Cos 30°= 20√3 m/sVertical component of the projectile = 40 m/s × Sin 30°= 40 m/s ×½ = 20m/s.Let us take the point of projection as the origin of coordinate system, upward direction as positive, and downward direction as negative, and the instant of firing at t= 0. The time of flight is the time from the time of firing to the time when it again returns to the ground. The time of flight of the projectile can be obtained using ,the relation,s = u× t + ½ a t². In this case s= 0 m, as the projectile returns to the ground, u = initial upwards velocity = 20 m/s, a = - 10 m/s², we have0 = 20 t - 5 t²; 5 t² - 20 t=0,==> 5t (t -4)=0 ; ==> t=0 or t= 4 s. Displacement was zero at start and again after 4 s.So range of the projectile= horizontal component × time of flight = 20√3 m/s × 4 s = 80√3 m = 138.56 m.
Projectile motion problem?
the range equation is R=v0^2 sin(2x)/g where x is the angle if v0 stays the same, we need the value of sin(2x) to double for x=12 degrees, sin(2x)=sin 24=0.41 to double the range, we need the value of sin 2x to equal 0.82 or we need 2x=arcsin(0.82)=54.4 deg so that x needs to be increased to 27.2 degrees