Find the speed at which the ball was launched. (Give your answer to two decimal

Question

Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.) 18 1285 m/s ocean Find the vertical distance by which the ball clears the wall 0.34 m Find the horizontal distance from the wall to the point on the roof where the ball lands. 2.4 m A car is parked on a diff overlooking the on an incline that makes an angle of 16.0 degree below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 2.59 m/s for a distance of 65.0 m to the edge of the diff, which is 35.0 m above the ocean. Find the car's position relative to the base of the diff when the car lands in the ocean. _____________ m Find the length of time the car is in the ___________ s A fireman d = 32.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of theta_t = 39.0 degree above the horizontal as shown in the figure If the speed of the stream as it leaves the hose is v_1 = 40.0 m/s, at what height will the stream of water strike the building? 20.7 m

Explanation / Answer

4.

Using the equation :vf2= vi2+2as

vf is the final speed ; vi is the initial speed

vi=0

a is the acceleration ,a= 2.59 m/s2

s = distance = 65 m

So the speed while leaving the cliff is :

vf2= 0+2(2.59)(65)= 336.7

vf= 18.35 m/s

Components of car's velocity when it leaves the cliff:

Vhorizontal= (18.35)cos(16°) = 17.64 m/s

Vvertical= -(18.35)sin(16°) = -5.06 m/s

Time , the car is in air can be determined as:

y(t)= y0+v0yt-(1/2)gt2

y(t) is the height at any time t

y0 is the initial position

V0y is the initial speed

How long it takes car to reach y=0 is given by:

0= 35-5.06t-(9.81/2)t2

4.9t2+5.06t-35 = 0

Solving this quadratic equation , we get

t= -3.24 & t=2.21 .

As time can not be negative so we choose

t= 2.21 s

Since horizontal speed will not change once the car leaves the cliff.

Hence horizontal distance traveled in 2.21 s:

x = vhorizontal×2.21s

= 17.64 m/s × 2.21s = 38.98 m

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