Question

a projectile is fired horizontally with a velocity of 48 m/s from the hill 294 m high. find the time taken to reach the ground.
a 10.0 s
b 7.75 s
c 15.2 s
d 60.5 s

Explanation:

Step1: Identify vertical - motion formula

The vertical - displacement of the projectile is given by the equation $h = v_{0y}t+\frac{1}{2}gt^{2}$. Since the projectile is fired horizontally, $v_{0y}=0$. So the equation simplifies to $h=\frac{1}{2}gt^{2}$.

Step2: Solve for time $t$

We can re - arrange the equation $h=\frac{1}{2}gt^{2}$ to solve for $t$. First, multiply both sides by 2 to get $2h = gt^{2}$. Then, $t=\sqrt{\frac{2h}{g}}$. Given $h = 294$ m and $g = 9.8$ m/s², we substitute these values: $t=\sqrt{\frac{2\times294}{9.8}}$.

Step3: Calculate the value of $t$

$t=\sqrt{\frac{588}{9.8}}=\sqrt{60}\approx 7.75$ s.

Answer:

b. 7.75 s