Kinematics: Question 8

Worked solution

Setting up the problem

The horizontal and vertical motions of the ball are independent: the horizontal component of velocity stays constant (no horizontal force, air resistance negligible), while the vertical component is uniformly decelerated (then accelerated) by gravity, $g=9.81\text{ m s}^{-2}$ downwards.

Part (a): Components of the initial velocity

Resolving the initial velocity $u=18.0\text{ m s}^{-1}$ at $35.0°$ above the horizontal: $u_x = u\cos35.0° = 18.0 \times 0.8192 = 14.7\text{ m s}^{-1} (3\text{ s.f.})$ $u_y = u\sin35.0° = 18.0 \times 0.5736 = 10.3\text{ m s}^{-1} (3\text{ s.f.})$

Part (b): Time to reach maximum height

At maximum height the vertical velocity is momentarily zero. Taking upwards as positive, $a=-g=-9.81\text{ m s}^{-2}$. Using $v_y = u_y + at$ with $v_y=0$: $0 = 10.324 - 9.81\,t_{up}$ $t_{up} = \frac{10.324}{9.81} = 1.05\text{ s} (3\text{ s.f.})$

Part (c): Maximum height reached

Using $v_y^2 = u_y^2 + 2ah_{max}$ with $v_y=0$ and $a=-9.81\text{ m s}^{-2}$: $0 = (10.324)^2 - 2(9.81)h_{max}$ $h_{max} = \frac{106.59}{19.62}$ $h_{max} = 5.43\text{ m} (3\text{ s.f.})$

Part (d): Total time of flight

Since the ball lands back at the same height (ground level) from which it was launched, the time to fall from maximum height back to the ground equals the time taken to rise to maximum height. So the total time of flight is: $T = 2\,t_{up} = 2(1.0524)$ $T = 2.10\text{ s} (3\text{ s.f.})$

Part (e): Horizontal range

The horizontal component of velocity is constant throughout the flight, so: $x = u_x T = 14.745 \times 2.1049$ $x = 31.0\text{ m} (3\text{ s.f.})$

Check: using the range formula $R=\dfrac{u^2\sin(2\theta)}{g}=\dfrac{(18.0)^2\sin70.0°}{9.81}=\dfrac{324\times0.9397}{9.81}=31.0\text{ m}$, which agrees.

Final answers

• (a) $u_x = \boxed{14.7}\text{ m s}^{-1}$, $u_y = \boxed{10.3}\text{ m s}^{-1}$
• (b) Time to maximum height $= \boxed{1.05}\text{ s}$
• (c) Maximum height $= \boxed{5.43}\text{ m}$
• (d) Total time of flight $= \boxed{2.10}\text{ s}$
• (e) Horizontal range $= \boxed{31.0}\text{ m}$