A blast blows a rock straight up at 150 ft/sec. The path of the rock is described by the s(t) = 150t-16t^2 where s is measured in feet and t in seconds. Find the velocity of the rock when it is 200 ft above the ground and when the rock hits the ground.
v(t)=s'(t)=150-32t
set s(t)=200 find out when the rock is 200 feet above the ground.
200=150t-16t^2
t=(5/16)+or-(15-sqrt(97))
0=150t-16t^2
t=75/8 not 0 because it started on the ground
the velocity when it strikes the ground is v(75/8)=-150
the other velocity is v((5/16)+or-(15-sqrt(97)))=304.387 or -24.836
To find the velocity of the rock at a given height and when it hits the ground, we need to differentiate the equation for its position with respect to time.
Given that the equation for the rock's position is s(t) = 150t - 16t^2, we can differentiate it to find the velocity function, v(t), which describes the rate of change of the position with respect to time.
The derivative of s(t) with respect to t is obtained by applying the power rule of differentiation:
ds/dt = d(150t - 16t^2)/dt
= 150 - 32t
Therefore, the velocity function, v(t), is given by v(t) = 150 - 32t.
To find the velocity when the rock is 200 ft above the ground, we substitute s(t) = 200 into the position equation and solve for t:
200 = 150t - 16t^2
This equation is a quadratic equation, so we can rearrange it to standard form:
16t^2 - 150t + 200 = 0
Using the quadratic formula, t can be calculated:
t = (-(-150) ± √((-150)^2 - 4(16)(200)))/(2(16))
t ≈ 4.16 seconds or t ≈ 7.59 seconds
Now, we substitute these values of t into the velocity function, v(t), to find the velocities:
v(4.16) = 150 - 32(4.16) ≈ 20.88 ft/sec
v(7.59) = 150 - 32(7.59) ≈ -60.88 ft/sec
Hence, when the rock is 200 ft above the ground, its velocity is approximately 20.88 ft/sec, and when it hits the ground, its velocity is approximately -60.88 ft/sec.