A projectile motion launched with velocity of 15 meter/s at an angle of 40 degre

Question

A projectile motion launched with velocity of 15 meter/s at an angle of 40 degrees with respect to the x-axis.
(a.) Determine the time it takes to reach its highest point.
(b.) the maximum height it reaches.
(c.) the time it takes to return to the same level as it was launched
(d.) its acceleration at highest point
(e.) its range.
(f.) what is the velocity of the projectile at time = 0.3 seconds.

please solve and explain it one step at a time. also space in between lines so that i will not confuse.

Explanation / Answer

The initial speed of the projectile is

u = 15 m/s

The angle of projection = 40

a)

Time taken to reach the heighest point is

Ta = usin/g

Ta = 15*sin40 / 9.8

Ta = 0.984 s

b)

The maximum height reached

H = u^2 (sin)^2 / 2g

H = (15sin40)^2 / 2*9.8

H = 4.743 m

c)

The time it takes to return to the same level as it was launched is

Td = usin/g

Td = 15*sin40 / 9.8

Td = 0.984 s

d)

At the heighest point of the trajectory, the projectile will move at constant horizontal speed so that there will be no acceleration in horizontal direction

But the projectile always experience acceleration due to gravity, downwards, through out the path

e)

The range is

R = u^2sin2/g

R = 15^2*sin80/9.8

R = 22.61 m

f)

The equations of motion of the projectile

in X direction:

vx = ucos + 0*t

vx = ucos

vx = 11.49 m/s

In Y direction:

vy = usin ± gt

At t = 0.3 s

t<Ta

vy = 15sin40 - 9.8*0.3

vy = 6.7 m/s

The velocity of the particle at t = 0.3 s is

v = (11.49^2 + 6.7^2)

v = 13.3 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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