can some one help me solve the math problem

Question

 Suppose that an arrow was fired vertically upward from a crossbow at ground level and that it struck the ground 20 seconds later.  If air resistance may be neglected, find the initial velocity of the bolt and the maxim altitude that it reached.

Answer 1

Using feet, ft/s, and g = 32 ft/s², the equation of motion is

h(t) = -(1/2)gt² + v₀t

h(t) = -16t² + v₀t

h(20) = 0 = -16(20)² +20v₀

v₀ = 320 ft/s

 

h(t) = -16t² + 320t

 

dh/dt = -32t + 320 = 0 ⇒ t = 320/32 s = 10 s

 

h_max is reached at t = 10 s

 

h_max = h(10) = [-16(10)² +320(10)] feet = 1600 feet

Answer 2

Assumptions: 1) The arrow starts at exactly ground level: H₀ = 0. 2)The arrow's motion is perfectly vertical with no horizontal component to the motion. 3)As stated in the problem, there is no air resistance. 4)In our coordinate system downward motion is negative and upward motion is positive. 

 

The equation for determining the distance can be determined from calculus.

 

The arrow starts off accelerating upwards downwards at G = 9.8 m/s₂ as soon as it is released. To find the velocity as a function of time we must integrate the acceleration function.

 

A(t) = -G -> V(t) = ∫(-G)dt = -∫G dt = -Gt + C, the constant will be the velocity at time 0 or the initial velocity.

 

V_o = -G(0) + C = C.

 

Thus, the function for Velocity given the time is: V(t) = -Gt + V_o

 

The equation for determining the arrow's height is found by integrating one step further.

 

H(t) = ∫V(t) dt = ∫(-Gt + Vo) dt = -(1/2)Gt² + V_ot; the constant is the initial height which is assumed zero.

 

 

The problem asks for the height of the maximum height of the arrow and the initial velocity.

 

A)Find Initial Velocity:

 

The height of the arrow is zero at two points, t = 0 and t = 20.

 

First, solve for V_o in height equation: V_o = (H(t) + (1/2)Gt²)/t

 

Setting t =20, (1/2)*(9.8 m/s²)*(20s)²/20s = 98 m/s.

 

initial velocity, V_o = 98 m/s.

 

B)Find maximum height.

 

The maximum height will be reached when velocity reaches zero.

 

Setting velocity to zero in the velocity equation:

 

0 = V_o - Gt -> V_o = Gt, t = V_o/G -> (98 m/s) / (9.8 m/s²) = 10 s.

 

The arrow reaches it maximum height at 10s.

 

Thus, the maximum height is H = 98 m/s * 10 s - (1/2) * 9.8 m/s² * (10 s)² = 490 m.

 

H_max = 490 m.