Question

A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.

Answer 1

Answer:

solved

Explanation:

x = 50tcos∝ , t = x sec∝/50

y = 50tsin∝ - 5t² + 40

y = xtan∝ -  x²sec²∝/500 + 40

at y = 0 , x = Range , t = total time

50tsin∝ - 5t² + 40 = 0

t² - 10tsin∝ - 8 = 0

t = 10sin∝±√100sin²∝ + 32 ÷ 2

t = 5sin∝±√25sin²∝ + 8 , the time should be positive → one root

t = 5sin∝+√25sin²∝ + 8      ---------------------------------(1)

vx = 50cos∝

vy = 50sin∝ - 10t

speed when it hits the ground is v

v² = vx² + vy² = 2500cos²∝ + 2500sin²∝ - 500tsin∝ + 100t²

v  = √(2500 - 500tsin∝+100t²)

t can be substituted from equation (1)

Answer 2

A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. The speed when it hits the ground is about 57.31 m/s

Explanation:

From the given statements, we know

Initial Velocity (u) = 50 m/s

Final Velocity       = v m/s

Height (h)             = 40 m

As per the Law of Conservation of Energy

Initial Total Energy = Final Total Energy

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

$\frac{1}{2} mu^{2} + mgh = \frac{1}{2} mv^{2} + mgh$

$( \frac{1}{2} \times m \times 2500) + m \times 9.8 \times 40 = (\frac{1}{2} \times m \times v^{2} ) + 0$

 [Final P.E. = 0 as final h= 0]

$1250 m + 392 m = \frac{1}{2} \times m \times v^{2}$

$1642 m = \frac{1}{2} \times m \times v^{2}$

$v^2 = 1642 \times 2 = 3284$

v = 57.306 m/s

Hence, the speed of the projectile when it hits the ground is 57.306 m/s or 57.31 m/s.