i) A projectile is shot from the edge of a cliff 10 m above the ground level wit

Question

i) A projectile is shot from the edge of a cliff 10 m above the ground level with initial speeds of20.0 ms, 25.0 m's, and 30.0 m's at an angle of 45.0° with the horizontal. ) Plot on the same graph, the graphs of yO vw x) for 0 tsSs for each of the initial speeds given. You may use time steps of 0.1 s b) For each initial speed, determine the maximum height (HD of the projectile above the ground level. c) For each initial speed, determine the range (R) of the projectile d) For each initial speed, determine the total time of flight (T) of the projectile. Note: Please attach a graph showing the trajectories of the projectile

Explanation / Answer

Let's assume the initial speed of projectile = u at an angle of = 45o with horizontal

Height of Cliff = 10m above ground

Vertical component of projectile speed = u sin45o

Horizontal component of projectile speed = u cos45o

Time taken for reaching projectile to ground can be calculated using vertical speed (u sin 45o), height (-10m; 10m below projectile initial positon) and acceleration due to gravity (g) as follows:

As we know that s = u*t +(1/2) * a *t2 ,

We must use a = -g (as acceleration due to gravity is opposite direction in this event

So, -10 = u sin45o * t + 0.5 *(-g) * t2

-0.5 * g* t2 + (u/2) * t +10 = 0

We can find value of "t" using quadratic equation:

t = { -(u/2) - (u2 /2 - 4* 0.5 *(- g) *10)}/(2*0.5*(-g)) (we are not considering other value of solution as that would be negative)

t = (1/g) * { (u2 /2 + 20 * g) + (u/2)}

Now, we can calculate time taken for different initial speed as 20, 25 & 30 m/s using g=9.8m/s2 and value of us as per initial speed value as follows

t20 = 3.47 sec (for u = 20m/s)

t25 = 4.10 sec (for u = 25m/s)

t30 = 4.76 sec (for u = 30m/s)

Above is the answer of part d)

Range of the projectile can be calculated by using Horizontal speed and time. as there is no acceleration in horizotal speed; Range = (Horizontal speed * time of projectile)

R20 = (u sin45o) * t20 = 20 * (1/2) * 3.47 = 49.13 m

R25 = (u sin45o) * t25 = 25 * (1/2) * 4.10 = 72.56 m

R30 = (u sin45o) * t30 = 30 * (1/2) * 4.76 = 100.94 m

Above is the answer for part c)

Maximum height of projectile can be calculated using formula v2 = u2 + 2*a*s, for vertical motion. ( s = (v2 - u2) /2a )

For maximum height, v = 0, u = u sin45o and a = -g = -9.8m/s2

s20 = ( 0 - (20 sin45o)2) /(2* (-9.8)) = 10.2 m

s25 = ( 0 - (25 sin45o)2) /(2* (-9.8)) = 15.94 m

s30 = ( 0 - (30 sin45o)2) /(2* (-9.8)) = 22.96 m

Now, we can calculate height of projectile above ground by adding height of cliff (10 m) in their respective "s" values

h20 = s20 + 10 = 10.2 + 10 =20.2 m

h25 = s25 + 10 = 15.94 + 10 =25.94 m

h30 = s30 + 10 = 22.96 + 10 = 32.96 m

Above is the answer of part b)

For part a) answer, meaning of y(t) and x(t) is not clear. If you have any further details for y(t) and x(t) , you may find using formula of "t" calculated above.

Horizontal movement does not have acceleration and distance can be calculated using Horizontal speed * Time

Vertical distance can be calculated using s = u* t + (1/2) * a * t2 , use a = -g, initial vertical speed = u sin45o,

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.