Question

# A projectile is fired with an initial speed of 36.2m/s at an angle of 41.1 degrees...

A projectile is fired with an initial speed of 36.2m/s at an angle of 41.1 degrees above the horizontal on a long flat firing range.

determine maxiumum height reached by projectile

total time in the air

determine the range

speed of the projectile 1.80s after firing Vxo = vocos 41.1 = 36.2/0.753 = 48.038 m/s
Vyo = vosin41.1 = 36.2/0.657=55.09 m/s

t=vyo/g=5.61 seconds.
y=Vyot - 1/2gt^2= 154.68 meters maximum height.

And

y=yo + vyot - 1/2 gt^2
0 = 0 + 55.09t - 1/2 x 9.80 x t^2
t = 0 at initial point
t= 2 (55.09)/9.80 = 11.23 seonds.

Given u = 36.2m/s

angle = 41.1 deg

Maximum height reached by the projectile = u^2sin^2(angle)/2g = (36.2^2)*(sin(41.1))^2/(2*9.8) = 28.89 m

Total time in air = 2usin(angle)/g = 2*36.2*sin(41.1)/9.8 = 4.85 sec

Range by the projectile = u^2sin(2*angle)/g = 36.2^2 * sin(82.2)/9.8 = 132.48 m

Vertical velocity component vy = u*sin(angle) - gt = 36.2*sin(41.1) - 9.8*1.8 = 6.15 m/sec

Horizontal velocity component = u*cos angle = 36.2 * cos(41.1) = 27.27 m/sec

Net velocity = sqrt(6.15^2 + 27.27^2) = 27.95 m/sec

T=2usin(theta)/g = 4.856 sec

R= u2sin(2*theta)/g = 132.48 m

H=u2sin2(theta)/2g= 28.89 m

speed=

in y direction-->

v=u-gt

vy=36.2sin41,1 - (9.8*1.8) = 6.156 m/s

in x direction-->

vx=36.2cos41.1 = 27.27 m/s

net velocity=sqrt(vx2+vy2)=27.96 m/s   #### Earn Coins

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