Question

Could you please solve it with shiwing the full work

1-

A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.

Take gravitational acceleration to be 9.81 m/s2.

2-

A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?

Take gravitational acceleration to be 9.81 m/s2.



3-

A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.

4-

A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.

Answer 1

1.V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

Step-by-step explanation:

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5 = 49 m/s : variable velocity in y

`v=\sqrt{v_(x) ^(2) +v_(y) ^(2) }`

`v=\sqrt{ 638.6^(2) +49 ^(2) }`

V= 640.48 m/s : total velocity in t= 5s

2.
`v_(ox) =v_(o) cos33.2=20.9*cos33,2= 17.49 m/s`

`v_(oy)=v_(o)*sin33,2 =20.9*sin33,2=11.44 m/s`

x=v₀x*t

13=v₀x*t

13=17.49*t

t=13/17.49=0.743s : time for 13.0 m away

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4. a = (-9t) m/s2

a=dv/dt=-9t

dv=-9tdt

v=∫ -9tdt

v=-9t²/2 + C1 equation (1)

in t=0 , v₀=26m/s ,in the equation (1) C1= 26

v=-9t²/2 + 26=ds/dt

ds=( -9t²/2 + 26)dt

s= ∫( -9t²/2 + 26)dt

s= -9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (-9t³/6+26t ) m

s= (-1.5t³+26t ) m